Daftar isi
Soal dan Pembahasan Integral Fungsi Aljabar adalah topik yang akan kita bahas kali ini. Topik pembahasan disesuaikan dengan kurikulum 2013 revisi yang merupakan pelajaran matematika wajib kelas 11 SMA. Berbeda dengan kurikulum 2006, materi integral pada kurikulum 2013 revisi telah mengalami penyempitan ruang lingkup, lebih sederhana dan lebih mudah untuk dipahami karena hanya sebatas integral fungsi aljabar biasa. Sebelum masuk ke topik utama yaitu Soal dan Pembahasan Integral Fungs Aljabar, kita akan melakukan review singkat tentang Integral Fungsi Aljabar.
INTEGRAL TAK TENTU
Pengertian Integral Tak Tentu
Integral adalah operasi invers (operasi kebalikan) dari diferensial (turunan). Integral disebut juga anti turunan (antiderivative) yang dinotasikan dengan $\displaystyle \int{f(x)dx}$ (artinya: integral $f(x)$ terhadap $x$). Dengan demikian $F(x)$ adalah integral dari $f(x)$ apabila $F'(x) = f(x)$ dan $\displaystyle \int{f(x)dx} = F(x) + C$. Integral sebagai operasi kebalikan (invers) dari turunan disebut juga sebagai integral tak tentu.Sifat-sifat Integral Tak Tentu
$1.\ \displaystyle \int{kf(x)\ dx} = k\displaystyle \int{f(x)\ dx}$
$2.\ \displaystyle \left[\int{f(x)} + \int{g(x)}\right]dx = \displaystyle \int{f(x)dx} + \displaystyle \int{g(x)}dx$
$3.\ \displaystyle \left[\int{f(x)} - \int{g(x)}\right]dx = \displaystyle \int{f(x)dx} - \displaystyle \int{g(x)}dx$
$2.\ \displaystyle \left[\int{f(x)} + \int{g(x)}\right]dx = \displaystyle \int{f(x)dx} + \displaystyle \int{g(x)}dx$
$3.\ \displaystyle \left[\int{f(x)} - \int{g(x)}\right]dx = \displaystyle \int{f(x)dx} - \displaystyle \int{g(x)}dx$
Rumus dan Contoh Soal Integral Tak Tentu
$1.\ \displaystyle \int{a\ dx} = ax + C$
$2.\ \displaystyle \int{ax^ndx} = \dfrac{a}{n + 1}x^{n + 1} + C$ dengan $n \ne -1$
$3.\ \displaystyle \int{\dfrac{a}{x}\ dx} = a\ ln\ |x| + C$ dimana $ln\ x =\ ^elog\ x$
$2.\ \displaystyle \int{ax^ndx} = \dfrac{a}{n + 1}x^{n + 1} + C$ dengan $n \ne -1$
$3.\ \displaystyle \int{\dfrac{a}{x}\ dx} = a\ ln\ |x| + C$ dimana $ln\ x =\ ^elog\ x$
Contoh soal 1.
Tentukanlah hasil dari:
$\displaystyle \int{4x^2\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{4x^2\ dx}$
$\displaystyle \int{ax^n\ dx} = \dfrac{a}{n + 1}x^{n + 1} + C$ dengan $n \ne -1$
$\displaystyle \int{4x^2\ dx} = \dfrac{4}{2 + 1}x^{2 + 1} + C$
$= \dfrac43x^3 + C$
$\displaystyle \int{4x^2\ dx} = \dfrac{4}{2 + 1}x^{2 + 1} + C$
$= \dfrac43x^3 + C$
Contoh soal 2.
Tentukanlah hasil dari:
$\displaystyle \int{5x^7\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{5x^7\ dx}$
$\displaystyle \int{5x^7\ dx} = \dfrac{5}{7 + 1}x^{7 + 1} + C$
$= \dfrac{5}{8}x^8 + C$
$= \dfrac{5}{8}x^8 + C$
Contoh soal 3.
Tentukanlah hasil dari:
$\displaystyle \int{2x^{-3}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{2x^{-3}\ dx}$
$\displaystyle \int{2x^{-5}\ dx} = \dfrac{2}{-5 + 1}x^{-5 + 1}$
$= \dfrac{2}{-4}x^{-4} + C$
$= -\dfrac{1}{2x^4} + C$
$= \dfrac{2}{-4}x^{-4} + C$
$= -\dfrac{1}{2x^4} + C$
Contoh soal 4.
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac32x^{-2}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac32x^{-2}\ dx}$
$\displaystyle \int{\dfrac32x^{-2}\ dx} = \dfrac32.\dfrac{1}{-2 + 1}x^{-2 + 1} + C$
$= \dfrac32.\dfrac{1}{-1}x^{-1} + C$
$= -\dfrac{3}{2x} + C$
$= \dfrac32.\dfrac{1}{-1}x^{-1} + C$
$= -\dfrac{3}{2x} + C$
Contoh soal 5.
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac43x^{3/2}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac43x^{3/2}\ dx}$
$\displaystyle \int{\dfrac43x^{3/2}\ dx} = \dfrac43.\dfrac{1}{\dfrac32 + 1}x^{3/2 + 1} + C$
$= \dfrac43.\dfrac25x^{5/2} + C$
$= \dfrac{8}{15}x^{5/2} + C$
$= \dfrac43.\dfrac25x^{5/2} + C$
$= \dfrac{8}{15}x^{5/2} + C$
Contoh soal 6.
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{2}{5x^2\sqrt{x}}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{2}{5x^2\sqrt{x}}\ dx}$
$\displaystyle \int{\dfrac{2}{5x^2\sqrt{x}}\ dx} = \displaystyle \int{\dfrac{2}{5}x^{-5/2}\ dx}$
$= \dfrac25.\dfrac{1}{-\dfrac52 + 1}x^{-5/2 + 1} + C$
$= \dfrac25.\left(-\dfrac{2}{3}\right)x^{-3/2} + C$
$= - \dfrac{4}{15}x^{-3/2} + C$
$= \dfrac25.\dfrac{1}{-\dfrac52 + 1}x^{-5/2 + 1} + C$
$= \dfrac25.\left(-\dfrac{2}{3}\right)x^{-3/2} + C$
$= - \dfrac{4}{15}x^{-3/2} + C$
Contoh soal 7.
Tentukanlah hasil dari:
$\displaystyle \int{(2x^5 + 3x^2 + 7)\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{(2x^5 + 3x^2 + 7)\ dx}$
$\displaystyle \int{(2x^5 + 3x^2 + 7)\ dx} = \dfrac{2}{5 + 1}x^{5 + 1} + \dfrac{3}{2 + 1}x^{2 + 1} + 7x + C$
$= \dfrac13x^6 + x^3 + 7x + C$
$= \dfrac13x^6 + x^3 + 7x + C$
Contoh soal 8.
Tentukanlah hasil dari:
$\displaystyle \int{(5x^3 - 3x\sqrt{x})\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{(5x^3 - 3x\sqrt{x})\ dx}$
$\displaystyle \int{(5x^3 - 3x\sqrt{x})\ dx} = \displaystyle \int{(5x^3 - 3x^{3/2})\ dx}$
$= \dfrac{5}{3 + 1}x^{3 + 1} - \dfrac{3}{\dfrac32 + 1}x^{3/2 + 1} + C$
$= \dfrac54x^4 - \dfrac65x^{5/2} + C$
$= \dfrac{5}{3 + 1}x^{3 + 1} - \dfrac{3}{\dfrac32 + 1}x^{3/2 + 1} + C$
$= \dfrac54x^4 - \dfrac65x^{5/2} + C$
Contoh soal 9.
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{4x^5 - 3x^3 + x}{x^2}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{4x^5 - 3x^3 + x}{x^2}\ dx}$
$\displaystyle \int{\dfrac{4x^5 - 3x^3 + x}{x^2}\ dx} = \displaystyle \int{(4x^3 - 3x + \dfrac 1x)\ dx}$
$= x^4 - \dfrac32x^2 + ln\ |x| + C$
$= x^4 - \dfrac32x^2 + ln\ |x| + C$
Contoh soal 10.
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{x^3 - 8}{x - 2}\ dx}$
Tentukanlah hasil dari:
$\displaystyle \int{\dfrac{x^3 - 8}{x - 2}\ dx}$
Dengan metode Horner kita bisa mencari hasil bagi dari $(x^3 - 8) : (x - 2)$
$\displaystyle \int{\dfrac{x^3 - 8}{x - 2}\ dx} = \displaystyle \int{\dfrac{(x - 2)(x^2 + 2x + 4)}{(x - 2)}\ dx}$
$= \displaystyle \int{(x^2 + 2x + 4)\ dx}$
$= \dfrac13x^3 + x^2 + 4x + C$
$\displaystyle \int{\dfrac{x^3 - 8}{x - 2}\ dx} = \displaystyle \int{\dfrac{(x - 2)(x^2 + 2x + 4)}{(x - 2)}\ dx}$
$= \displaystyle \int{(x^2 + 2x + 4)\ dx}$
$= \dfrac13x^3 + x^2 + 4x + C$
INTEGRAL TENTU
Pengertian Integral Tentu
Integral tentu adalah integral yang merupakan limit dari suatu jumlah. Integral tentu dinotasikan dengan $\displaystyle \int_{a}^{b}f(x)\ dx$ (artinya: integral fungsi $f(x)$ terhadap $x$ dengan batas $a$ sampai $b$). Fungsi $f(x)$ disebut integran dan $a$ dan $b$ disebut batas bawah dan batas atas pengintegralan. Jika $f(x)$ kontinu pada interval $a \leq x \leq b$ dan $F(x)$ merupakan antiderivative (anti turunan) dari $f(x)$ maka:Sifat-sifat Integral tentu
$1.\ \displaystyle \int_a^af(x)dx = 0$
$2.\ \displaystyle \int_a^bf(x)dx = -\displaystyle \int_b^af(x)dx$
$3.\ \displaystyle \int_a^bpf(x)dx = \displaystyle p\int_a^bf(x)dx$
$4.\ \displaystyle \int_a^b(f(x) \pm g(x))dx = \displaystyle \int_a^bf(x)dx \pm \displaystyle \int_a^bg(x)dx$
$5.\ \displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
$2.\ \displaystyle \int_a^bf(x)dx = -\displaystyle \int_b^af(x)dx$
$3.\ \displaystyle \int_a^bpf(x)dx = \displaystyle p\int_a^bf(x)dx$
$4.\ \displaystyle \int_a^b(f(x) \pm g(x))dx = \displaystyle \int_a^bf(x)dx \pm \displaystyle \int_a^bg(x)dx$
$5.\ \displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
Rumus dan Contoh Soal Integral Tentu
$\displaystyle \int_{a}^{b}f(x)\ dx = \left.\begin{matrix} F(x) \end{matrix}\right|_a^b = F(b) - F(a)$
Contoh soal 11.
Tentukanlah hasil dari:
$\displaystyle \int_{1}^{3}(3x^2 + 2x + 5) dx$
Tentukanlah hasil dari:
$\displaystyle \int_{1}^{3}(3x^2 + 2x + 5) dx$
$\displaystyle \int_{1}^{3}(3x^2 + 2x + 5) dx = \left.\begin{matrix} (x^3 + x^2 + 5x) \end{matrix}\right|_1^3$
$= 3^3 + 3^2 + 5.3 - (1^3 + 1^2 + 5.1)$
$= 27 + 9 + 15 - 7$
$= 44$
$= 3^3 + 3^2 + 5.3 - (1^3 + 1^2 + 5.1)$
$= 27 + 9 + 15 - 7$
$= 44$
Contoh soal 12.
Jika $n > 0$ dan $\displaystyle \int_{1}^{n}(2x - 3) dx = 12$, maka nilai $n$ adalah . . . .
Jika $n > 0$ dan $\displaystyle \int_{1}^{n}(2x - 3) dx = 12$, maka nilai $n$ adalah . . . .
$\displaystyle \int_{1}^{n}(2x - 3) dx = 12$
$\left.\begin{matrix} (x^2 - 3x) \end{matrix}\right|_1^n = 12$
$n^2 - 3n -(1^2 - 3.1) = 12$
$n^2 - 3n + 2 = 12$
$n^2 - 3n - 10 = 0$
$(n + 2)(n - 5) = 0$
$n = -2\ atau\ n = 5$
Karena $n > 0,\ maka\ n = 5$
$\left.\begin{matrix} (x^2 - 3x) \end{matrix}\right|_1^n = 12$
$n^2 - 3n -(1^2 - 3.1) = 12$
$n^2 - 3n + 2 = 12$
$n^2 - 3n - 10 = 0$
$(n + 2)(n - 5) = 0$
$n = -2\ atau\ n = 5$
Karena $n > 0,\ maka\ n = 5$
INTEGRAL DENGAN METODE SUBSTITUSI
I. Jika $\displaystyle \int F(x)\ dx = \displaystyle k\int (f(x))^n.f'(x)\ dx$, maka $f(x)$ bisa dimisalkan sebagai $u$ dan selanjutnya $du = f'(x)dx$.Contoh soal 13.
Tentukanlah hasil dari:
$\displaystyle \int x^2\sqrt{x^3 - 1}\ dx$
Tentukanlah hasil dari:
$\displaystyle \int x^2\sqrt{x^3 - 1}\ dx$
$\displaystyle \int x^2\sqrt{x^3 - 1}\ dx = \displaystyle \int \sqrt{x^3 - 1}.x^2dx$
Misalkan:
$u = x^3 - 1$
$\dfrac{du}{dx} = 3x^2$
$\dfrac13du = x^2dx$
Dengan demikian:
$\displaystyle \int \sqrt{x^3 - 1}.x^2dx = \displaystyle \int u^{1/2}.\dfrac13du$
$= \dfrac13\displaystyle \int u^{1/2}du$
$= \dfrac13.\dfrac{1}{\dfrac12 + 1}x^{1/2 + 1} + C$
$= \dfrac13.\dfrac23u^{3/2} + C$
$= \dfrac29u^{3/2} + C$
$= \dfrac29(x^3 - 1)^{3/2} + C$
Misalkan:
$u = x^3 - 1$
$\dfrac{du}{dx} = 3x^2$
$\dfrac13du = x^2dx$
Dengan demikian:
$\displaystyle \int \sqrt{x^3 - 1}.x^2dx = \displaystyle \int u^{1/2}.\dfrac13du$
$= \dfrac13\displaystyle \int u^{1/2}du$
$= \dfrac13.\dfrac{1}{\dfrac12 + 1}x^{1/2 + 1} + C$
$= \dfrac13.\dfrac23u^{3/2} + C$
$= \dfrac29u^{3/2} + C$
$= \dfrac29(x^3 - 1)^{3/2} + C$
Contoh soal 14.
Tentukanlah hasil dari:
$\displaystyle \int_0^1 \dfrac{x}{\sqrt{3x^2 + 1}}\ dx$
Tentukanlah hasil dari:
$\displaystyle \int_0^1 \dfrac{x}{\sqrt{3x^2 + 1}}\ dx$
$\displaystyle \int \dfrac{x}{\sqrt{3x^2 + 1}}\ dx = \displaystyle \int \dfrac{1}{\sqrt{3x^2 + 1}}.xdx$
Misalkan:
$u = 3x^2 + 1$
$\dfrac{du}{dx} = 6x$
$\dfrac16du = xdx$
Jika $x = 0 → u = 1$
Jika $x = 1 → u = 4$
Dengan demikian:
$\displaystyle \int_0^1 \dfrac{1}{\sqrt{3x^2 + 1}}.xdx = \displaystyle \int_1^4 \dfrac{1}{\sqrt{u}}.\dfrac16du$
$= \dfrac16\displaystyle \int_1^4 \dfrac{1}{\sqrt{u}}du$
$= \dfrac16\displaystyle \int_1^4 u^{-1/2}du$
$= \left.\begin{matrix}\dfrac16.\dfrac{1}{-\dfrac12 + 1}u^{-1/2 + 1}\end{matrix}\right|_1^4$
$= \left.\begin{matrix}\dfrac16.2.u^{1/2}\end{matrix}\right|_1^4$
$= \left.\begin{matrix}\dfrac13u^{1/2}\end{matrix}\right|_1^4$
$= \dfrac13(\sqrt{4} - \sqrt{1})$
$= \dfrac13(2 - 1)$
$= \dfrac13$
Misalkan:
$u = 3x^2 + 1$
$\dfrac{du}{dx} = 6x$
$\dfrac16du = xdx$
Jika $x = 0 → u = 1$
Jika $x = 1 → u = 4$
Dengan demikian:
$\displaystyle \int_0^1 \dfrac{1}{\sqrt{3x^2 + 1}}.xdx = \displaystyle \int_1^4 \dfrac{1}{\sqrt{u}}.\dfrac16du$
$= \dfrac16\displaystyle \int_1^4 \dfrac{1}{\sqrt{u}}du$
$= \dfrac16\displaystyle \int_1^4 u^{-1/2}du$
$= \left.\begin{matrix}\dfrac16.\dfrac{1}{-\dfrac12 + 1}u^{-1/2 + 1}\end{matrix}\right|_1^4$
$= \left.\begin{matrix}\dfrac16.2.u^{1/2}\end{matrix}\right|_1^4$
$= \left.\begin{matrix}\dfrac13u^{1/2}\end{matrix}\right|_1^4$
$= \dfrac13(\sqrt{4} - \sqrt{1})$
$= \dfrac13(2 - 1)$
$= \dfrac13$
II. Mengintegralkan bentuk $\sqrt[n]{ax + b}$
Untuk mengintegralkan bentuk $\sqrt[n]{ax + b}$ lakukan pemisalan $y = \sqrt[n]{ax + b}$ dan selanjutnya rubah integran ke dalam bentuk $f(y)$.
Contoh soal 15.
Tentukanlah hasil dari:
$\displaystyle \int x\sqrt[3]{x + 2}\ dx$
Tentukanlah hasil dari:
$\displaystyle \int x\sqrt[3]{x + 2}\ dx$
$\displaystyle \int x\sqrt[3]{x + 2}\ dx$
Misalkan:
$y = \sqrt[3]{x + 2} → y^3 = x + 2 → x = y^3 - 2$
Turunan bentuk implisit
$y^3 = x + 2 → 3y^2dy = dx$
Dengan demikian:
$\displaystyle \int x\sqrt[3]{x + 2}\ dx = \displaystyle \int (y^3 - 2).y.3y^2dy$
$= \displaystyle \int (3y^6 - 6y^3)dy$
$= \dfrac37y^7 - \dfrac32y^4 + C$
$= \dfrac37(\sqrt[3]{x + 2})^7 - \dfrac32(\sqrt[3]{x + 2})^4 + C$
$= \dfrac37(x + 2)^{7/3} - \dfrac32(x + 2)^{4/3} + C$
Misalkan:
$y = \sqrt[3]{x + 2} → y^3 = x + 2 → x = y^3 - 2$
Turunan bentuk implisit
$y^3 = x + 2 → 3y^2dy = dx$
Dengan demikian:
$\displaystyle \int x\sqrt[3]{x + 2}\ dx = \displaystyle \int (y^3 - 2).y.3y^2dy$
$= \displaystyle \int (3y^6 - 6y^3)dy$
$= \dfrac37y^7 - \dfrac32y^4 + C$
$= \dfrac37(\sqrt[3]{x + 2})^7 - \dfrac32(\sqrt[3]{x + 2})^4 + C$
$= \dfrac37(x + 2)^{7/3} - \dfrac32(x + 2)^{4/3} + C$
Contoh soal 16.
Tentukanlah hasil dari:
$\displaystyle \int \dfrac{1}{x + \sqrt{x}}\ dx$
Tentukanlah hasil dari:
$\displaystyle \int \dfrac{1}{x + \sqrt{x}}\ dx$
$\displaystyle \int \dfrac{1}{x + \sqrt{x}}\ dx = \displaystyle \int \dfrac{1}{\sqrt{x}(\sqrt{x} + 1)}\ dx$
$= \displaystyle \int \dfrac{1}{(\sqrt{x} + 1)}.\dfrac{dx}{\sqrt{x}}$
Misalkan:
$y = \sqrt{x} + 1 → \sqrt{x} = y - 1$
$y = \sqrt{x} + 1$
$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
$2dy = \dfrac{dx}{\sqrt{x}}$
Dengan demikian:
$= \displaystyle \int \dfrac{1}{(\sqrt{x} + 1)}.\dfrac{dx}{\sqrt{x}} = \displaystyle \int \dfrac{1}{y}.2dy$
$= 2\displaystyle \int \dfrac 1ydy$
$= 2\ ln\ |y| + C$
$= 2\ ln\ |\sqrt{x} + 1| + C$
$= \displaystyle \int \dfrac{1}{(\sqrt{x} + 1)}.\dfrac{dx}{\sqrt{x}}$
Misalkan:
$y = \sqrt{x} + 1 → \sqrt{x} = y - 1$
$y = \sqrt{x} + 1$
$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
$2dy = \dfrac{dx}{\sqrt{x}}$
Dengan demikian:
$= \displaystyle \int \dfrac{1}{(\sqrt{x} + 1)}.\dfrac{dx}{\sqrt{x}} = \displaystyle \int \dfrac{1}{y}.2dy$
$= 2\displaystyle \int \dfrac 1ydy$
$= 2\ ln\ |y| + C$
$= 2\ ln\ |\sqrt{x} + 1| + C$
Pembahasan Soal UNBK dan SBMPTN Integral Fungsi Aljabar
1. Soal Integral UN / UNBK MtkIPA 2019
$\displaystyle \int (3x^2 - 5x + 4)dx =$ . . . .
$A.\ x^3 - \dfrac52x^2 + 4x + C$
$B.\ x^3 - 5x^2 + 4x + C$
$C.\ 3x^3 - 5x^2 + 4x + C$
$D.\ 6x^3 - 5x^2 + 4x + C$
$E.\ 6x^3 - \dfrac52x^2 + 4x + C$
$A.\ x^3 - \dfrac52x^2 + 4x + C$
$B.\ x^3 - 5x^2 + 4x + C$
$C.\ 3x^3 - 5x^2 + 4x + C$
$D.\ 6x^3 - 5x^2 + 4x + C$
$E.\ 6x^3 - \dfrac52x^2 + 4x + C$
$\displaystyle \int (3x^2 - 5x + 4)dx = x^3 - \dfrac52x^2 + 4x + C$
jawab: A.
jawab: A.
2. Soal Integral UN / UNBK MtkIPA 2019
Hasil dari $\displaystyle \int (2x - 1)(x^2 - x + 3)^3dx =$ . . . .
$A.\ \dfrac13\left(x^2 - x + 3\right)^3 + C$
$B.\ \dfrac14\left(x^2 - x + 3\right)^3 + C$
$C.\ \dfrac14\left(x^2 - x + 3\right)^4 + C$
$D.\ \dfrac12\left(x^2 - x + 3\right)^4 + C$
$E.\ \left(x^2 - x + 3\right)^4 + C$
$A.\ \dfrac13\left(x^2 - x + 3\right)^3 + C$
$B.\ \dfrac14\left(x^2 - x + 3\right)^3 + C$
$C.\ \dfrac14\left(x^2 - x + 3\right)^4 + C$
$D.\ \dfrac12\left(x^2 - x + 3\right)^4 + C$
$E.\ \left(x^2 - x + 3\right)^4 + C$
Misalkan $u = x^2 - x + 3$
$\dfrac{du}{dx} = 2x - 1$
$du = (2x - 1)dx$
$\displaystyle \int (2x - 1)(x^2 - x + 3)dx = \int u^3du$
$= \dfrac14u^4 + C$
$= \dfrac14(x^2 - x + 3)^4 + C$
jawab: C.
$\dfrac{du}{dx} = 2x - 1$
$du = (2x - 1)dx$
$\displaystyle \int (2x - 1)(x^2 - x + 3)dx = \int u^3du$
$= \dfrac14u^4 + C$
$= \dfrac14(x^2 - x + 3)^4 + C$
jawab: C.
3. Soal Integral UN / UNBK MtkIPS 2019
Hasil dari $\displaystyle \int (2x^3 - 9x^2 + 4x - 5)dx =$ . . . .
$A.\ \dfrac12x^4 - 6x^3 + 2x^2 - 5x + C$
$B.\ \dfrac12x^4 - 6x^3 + x^2 - 5x + C$
$C.\ \dfrac12x^4 - 3x^3 + x^2 - 5x + C$
$D.\ \dfrac12x^4 - 3x^3 + 2x^2 - 5x + C$
$E.\ \dfrac12x^4 - 6x^3 - 2x^2 - 5x + C$
$A.\ \dfrac12x^4 - 6x^3 + 2x^2 - 5x + C$
$B.\ \dfrac12x^4 - 6x^3 + x^2 - 5x + C$
$C.\ \dfrac12x^4 - 3x^3 + x^2 - 5x + C$
$D.\ \dfrac12x^4 - 3x^3 + 2x^2 - 5x + C$
$E.\ \dfrac12x^4 - 6x^3 - 2x^2 - 5x + C$
$\displaystyle \int (2x^3 - 9x^2 + 4x - 5)dx$
$= \dfrac24x^4 - \dfrac93x^3 + \dfrac42x^2 - 5x + C$
$= \dfrac12x^4 - 3x^3 + 2x^2 - 5x + C$.
jawab: D.
$= \dfrac24x^4 - \dfrac93x^3 + \dfrac42x^2 - 5x + C$
$= \dfrac12x^4 - 3x^3 + 2x^2 - 5x + C$.
jawab: D.
4. Soal Integral UN MtkIPS 2018
$\displaystyle \int_{0}^{2}x(3x + 5)dx =$ . . . .
$A.\ 18$
$B.\ 16$
$C.\ 15$
$D.\ 10$
$E.\ 6$
$A.\ 18$
$B.\ 16$
$C.\ 15$
$D.\ 10$
$E.\ 6$
Ingat !
$\displaystyle \int ax^ndx = \frac{a}{n + 1}x^{n + 1} + c$
$\displaystyle \int_{0}^{2}x(3x + 5)dx$
$= \displaystyle \int_{0}^{2}(3x^2 + 5x)dx$
$= (x^3 + \dfrac{5}{2}x^2)|_{0}^{2}$
$= 2^3 + \dfrac{5}{2}.2^2 - (0^3 + \dfrac{5}{2}.0^2)$
$= 8 + 10$
$= 18.
jawab: A.
$\displaystyle \int ax^ndx = \frac{a}{n + 1}x^{n + 1} + c$
$\displaystyle \int_{0}^{2}x(3x + 5)dx$
$= \displaystyle \int_{0}^{2}(3x^2 + 5x)dx$
$= (x^3 + \dfrac{5}{2}x^2)|_{0}^{2}$
$= 2^3 + \dfrac{5}{2}.2^2 - (0^3 + \dfrac{5}{2}.0^2)$
$= 8 + 10$
$= 18.
jawab: A.
5. Soal Integral UN MtkIPS 2018
$\displaystyle \int_{-1}^{2}(6x^2 + 8x)dx =$ . . . .
$A.\ 24$
$B.\ 26$
$C.\ 28$
$D.\ 30$
$E.\ 32$
$A.\ 24$
$B.\ 26$
$C.\ 28$
$D.\ 30$
$E.\ 32$
$\displaystyle \int_{-1}^{2}(6x^2 + 8x)dx = (2x^3 + 4x^2)|_{-1}^{2}$
$= (2.2^3 + 4.2^2 - (2.(-1)^3 + 4(-1)^2)$
$= 16 + 16 - (-2 + 4)$
$= 30$
jawab: D.
$= (2.2^3 + 4.2^2 - (2.(-1)^3 + 4(-1)^2)$
$= 16 + 16 - (-2 + 4)$
$= 30$
jawab: D.
6. Soal Integral UN MtkIPA 2018
Hasil dari $\displaystyle \int \dfrac{x-1}{\sqrt{x^2-2x+10}}dx =$ . . . .
$A.\ -\sqrt{x^2-2x+10} + C$
$B.\ -\dfrac{1}{2}\sqrt{x^2-2x+10} + C$
$C.\ \dfrac{1}{2}\sqrt{x^2-2x+10} + C$
$D.\ \sqrt{x^2-2x+10} + C$
$E.\ 2\sqrt{x^2-2x+10} + C$
$A.\ -\sqrt{x^2-2x+10} + C$
$B.\ -\dfrac{1}{2}\sqrt{x^2-2x+10} + C$
$C.\ \dfrac{1}{2}\sqrt{x^2-2x+10} + C$
$D.\ \sqrt{x^2-2x+10} + C$
$E.\ 2\sqrt{x^2-2x+10} + C$
Misalkan:
$u = x^2 - 2x + 10$
$\displaystyle \dfrac{du}{dx} = 2x -2$
$\displaystyle \dfrac{du}{dx} = 2(x - 1)$
$\displaystyle \dfrac{1}{2}du = (x - 1)dx$
$\displaystyle \int \dfrac{x-1}{\sqrt{x^2-2x+10}}dx = \dfrac{1}{2}\displaystyle \int \dfrac{du}{\sqrt{u}}$
$\displaystyle = {1\over 2}\int u^{-1/2}du$
$\displaystyle = {1\over 2}.\dfrac{1}{-\dfrac12 + 1}.u^{-1/2 + 1} + C$
$\displaystyle = {1\over 2}.2.u^{1/2} + C$
$\displaystyle = \sqrt{u} + C$
$\displaystyle = \sqrt{x^2 -2x + 10} + C$
jawab: D.
$u = x^2 - 2x + 10$
$\displaystyle \dfrac{du}{dx} = 2x -2$
$\displaystyle \dfrac{du}{dx} = 2(x - 1)$
$\displaystyle \dfrac{1}{2}du = (x - 1)dx$
$\displaystyle \int \dfrac{x-1}{\sqrt{x^2-2x+10}}dx = \dfrac{1}{2}\displaystyle \int \dfrac{du}{\sqrt{u}}$
$\displaystyle = {1\over 2}\int u^{-1/2}du$
$\displaystyle = {1\over 2}.\dfrac{1}{-\dfrac12 + 1}.u^{-1/2 + 1} + C$
$\displaystyle = {1\over 2}.2.u^{1/2} + C$
$\displaystyle = \sqrt{u} + C$
$\displaystyle = \sqrt{x^2 -2x + 10} + C$
jawab: D.
7. Soal Integral UN MtkIPA 2018
Diketahui $\displaystyle \int_{0}^{3}\left(x^2 + px + 2 \right)dx = \left({3\over 2} \right )$. Nilai
$p$ yang memenuhi adalah . . . .
$A.\ -26$
$B.\ -13$
$C.\ -3$
$D.\ 3$
$E.\ 13$
$p$ yang memenuhi adalah . . . .
$A.\ -26$
$B.\ -13$
$C.\ -3$
$D.\ 3$
$E.\ 13$
$\displaystyle \int_{0}^{3}\left(x^2 + px + 2 \right)dx = \left(\dfrac{3}{2} \right )$
$\displaystyle \left[\dfrac{1}{3}x^3 + \dfrac{p}{2}x^2 + 2x \right]_{0}^{3} = \dfrac{3}{2}$
$\displaystyle \dfrac{1}{3}.3^3 + \dfrac{p}{2}.3^2 + 2.3 - 0 = \dfrac{3}{2}$
$\displaystyle 9 + \dfrac{9}{2}p + 6 = \dfrac{3}{2}$
$\displaystyle 15 + \dfrac{9}{2}p = \dfrac{3}{2}$
$30 + 9p = 3$
$9p = -27$
$p = -3$
jawab: C.
$\displaystyle \left[\dfrac{1}{3}x^3 + \dfrac{p}{2}x^2 + 2x \right]_{0}^{3} = \dfrac{3}{2}$
$\displaystyle \dfrac{1}{3}.3^3 + \dfrac{p}{2}.3^2 + 2.3 - 0 = \dfrac{3}{2}$
$\displaystyle 9 + \dfrac{9}{2}p + 6 = \dfrac{3}{2}$
$\displaystyle 15 + \dfrac{9}{2}p = \dfrac{3}{2}$
$30 + 9p = 3$
$9p = -27$
$p = -3$
jawab: C.
8. Soal Integral UN MtkIPA 2017
Hasil dari $\displaystyle \int \dfrac{x + 2}{\sqrt{x^2 + 4x - 3}}\ dx$ adalah . . . .
$A.\ \sqrt{x^2 + 4x - 3} + C$
$B.\ 2\sqrt{x^2 + 4x - 3} + C$
$C.\ 3\sqrt{x^2 + 4x - 3} + C$
$D.\ 4\sqrt{x^2 + 4x - 3} + C$
$E.\ 6\sqrt{x^2 + 4x - 3} + C$
$A.\ \sqrt{x^2 + 4x - 3} + C$
$B.\ 2\sqrt{x^2 + 4x - 3} + C$
$C.\ 3\sqrt{x^2 + 4x - 3} + C$
$D.\ 4\sqrt{x^2 + 4x - 3} + C$
$E.\ 6\sqrt{x^2 + 4x - 3} + C$
Misalkan:
$u = x^2 + 4x - 3$
$\dfrac{du}{dx} = 2x + 4$
$\dfrac{du}{dx} = 2(x + 2)$
$\dfrac12du = (x + 2)dx$
$\displaystyle \int \dfrac{x + 2}{\sqrt{x^2 + 4x - 3}}\ dx = \dfrac12\displaystyle \int \dfrac{1}{\sqrt{u}}\ du$
$= \displaystyle \dfrac12\int u^{-1/2}du$
$= \dfrac12.\dfrac{1}{-\dfrac12 + 1}u^{-1/2 + 1} + C$
$= \dfrac12.\dfrac{1}{\dfrac12}.u^{1/2} + C$
$= \dfrac12.2.u^{1/2} + C$
$= \sqrt{u} + C$
$= \sqrt{x^2 + 4x - 3} + C$
jawab: A.
$u = x^2 + 4x - 3$
$\dfrac{du}{dx} = 2x + 4$
$\dfrac{du}{dx} = 2(x + 2)$
$\dfrac12du = (x + 2)dx$
$\displaystyle \int \dfrac{x + 2}{\sqrt{x^2 + 4x - 3}}\ dx = \dfrac12\displaystyle \int \dfrac{1}{\sqrt{u}}\ du$
$= \displaystyle \dfrac12\int u^{-1/2}du$
$= \dfrac12.\dfrac{1}{-\dfrac12 + 1}u^{-1/2 + 1} + C$
$= \dfrac12.\dfrac{1}{\dfrac12}.u^{1/2} + C$
$= \dfrac12.2.u^{1/2} + C$
$= \sqrt{u} + C$
$= \sqrt{x^2 + 4x - 3} + C$
jawab: A.
9. Soal Integral UN MtkIPA 2017
Nilai $\displaystyle \int_2^4 (6x^2 - 6x - 1)\ dx$ adalah . . . .
A. 64
B. 68
C. 72
D. 74
E. 76
A. 64
B. 68
C. 72
D. 74
E. 76
$\displaystyle \int_2^4 (6x^2 - 6x - 1)\ dx = (2x^3 - 3x^2 - x)\bigr|_2^4$
$= 2.4^3 - 3.4^2 - 4 - (2.2^3 - 3.2^2 - 2)$
$= 2.64 - 3.16 - 4 - (2.8 - 3.4 - 2)$
$= 128 - 48 - 4 - (16 - 12 - 2)$
$= 76 - 2$
$= 74$
jawab: D.
$= 2.4^3 - 3.4^2 - 4 - (2.2^3 - 3.2^2 - 2)$
$= 2.64 - 3.16 - 4 - (2.8 - 3.4 - 2)$
$= 128 - 48 - 4 - (16 - 12 - 2)$
$= 76 - 2$
$= 74$
jawab: D.
10. Soal Integral UN MtkIPS 2017
Hasil dari $\displaystyle \int_{-1}^3 (6x^2 + 5)\ dx$ adalah . . . .
A. 103
B. 76
C. 62
D. 40
E. 26
A. 103
B. 76
C. 62
D. 40
E. 26
$\displaystyle \int_{-1}^3 (6x^2 + 5)\ dx = (2x^3 + 5x)\bigr|_{-1}^3$
$= 2.3^3 + 5.3 - (2.(-1)^3 + 5.(-1))$
$= 2.27 + 15 - (-2 - 5)$
$= 54 + 15 - (-7)$
$= 54 + 15 + 7$
$= 76$
jawab: B.
$= 2.3^3 + 5.3 - (2.(-1)^3 + 5.(-1))$
$= 2.27 + 15 - (-2 - 5)$
$= 54 + 15 - (-7)$
$= 54 + 15 + 7$
$= 76$
jawab: B.
11. Soal Integral UN MtkIPA 2016
Hasil $\displaystyle \int 2x(5 - x)^3\ dx = \cdots$
$A.\ -\dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
$B.\ -\dfrac{1}{10}(6x + 5)(5 - x)^4 + C$
$C.\ -\dfrac{1}{10}(x + 5)(5 - x)^4 + C$
$D.\ \dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
$E.\ \dfrac12(5 + x)^4 + C$
$A.\ -\dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
$B.\ -\dfrac{1}{10}(6x + 5)(5 - x)^4 + C$
$C.\ -\dfrac{1}{10}(x + 5)(5 - x)^4 + C$
$D.\ \dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
$E.\ \dfrac12(5 + x)^4 + C$
Misalkan:
$u = 5 - x → x = 5 - u$
$\dfrac{du}{dx} = -1 → dx = -du$
$\displaystyle \int 2x(5 - x)^3\ dx = \displaystyle -\int 2(5 - u)u^3\ du$
$= \displaystyle -\int 10u^3\ du + \displaystyle \int 2u^4\ du$
$= -\dfrac52u^4 + \dfrac25u^5 + C$
$= -\dfrac{25}{10}u^4 + \dfrac{4}{10}u^5 + C$
$= -\dfrac{1}{10}(25 - 4u)u^4 + C$
$= -\dfrac{1}{10}(25 - 4(5 - x))(5 - x)^4 + C$
$= -\dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
jawab: A.
$u = 5 - x → x = 5 - u$
$\dfrac{du}{dx} = -1 → dx = -du$
$\displaystyle \int 2x(5 - x)^3\ dx = \displaystyle -\int 2(5 - u)u^3\ du$
$= \displaystyle -\int 10u^3\ du + \displaystyle \int 2u^4\ du$
$= -\dfrac52u^4 + \dfrac25u^5 + C$
$= -\dfrac{25}{10}u^4 + \dfrac{4}{10}u^5 + C$
$= -\dfrac{1}{10}(25 - 4u)u^4 + C$
$= -\dfrac{1}{10}(25 - 4(5 - x))(5 - x)^4 + C$
$= -\dfrac{1}{10}(4x + 5)(5 - x)^4 + C$
jawab: A.
12. Soal Integral UN MtkIPA 2016
Nilai dari $\displaystyle \int_{-1}^1 (2x^2 - 4x + 3)\ dx = \cdots$
$A.\ \dfrac{22}{3}$
$B.\ 6$
$C.\ \dfrac{16}{3}$
$D.\ 4$
$E.\ \dfrac43$
$A.\ \dfrac{22}{3}$
$B.\ 6$
$C.\ \dfrac{16}{3}$
$D.\ 4$
$E.\ \dfrac43$
$\displaystyle \int_{-1}^1 (2x^2 - 4x + 3)\ dx = \left[\dfrac23x^3 - 2x^2 + 3x\right]_{-1}^1$
$= \dfrac23.1^3 - 2.1^2 + 3.1 - \left(\dfrac23.(-1)^3 - 2.(-1)^2 + 3.(-1)\right)$
$= \dfrac23 - 2 + 3 - \left(-\dfrac23 - 2 - 3\right)$
$= 1\dfrac23 + 5\dfrac23$
$= 7\dfrac13$
$= \dfrac{22}{3}$
jawab: A.
$= \dfrac23.1^3 - 2.1^2 + 3.1 - \left(\dfrac23.(-1)^3 - 2.(-1)^2 + 3.(-1)\right)$
$= \dfrac23 - 2 + 3 - \left(-\dfrac23 - 2 - 3\right)$
$= 1\dfrac23 + 5\dfrac23$
$= 7\dfrac13$
$= \dfrac{22}{3}$
jawab: A.
13. Soal Integral UN MtkIPA 2016
Hasil dari $\displaystyle \int sin^5\ 2xcos\ 2x\ dx = \cdots$
$A.\ -\dfrac15sin^62x + C$
$B.\ -\dfrac{1}{10}sin^62x + C$
$C.\ -\dfrac{1}{12}sin^62x + C$
$D.\ \dfrac{1}{12}sin^62x + C$
$E.\ \dfrac{1}{10}sin^62x + C$
$A.\ -\dfrac15sin^62x + C$
$B.\ -\dfrac{1}{10}sin^62x + C$
$C.\ -\dfrac{1}{12}sin^62x + C$
$D.\ \dfrac{1}{12}sin^62x + C$
$E.\ \dfrac{1}{10}sin^62x + C$
Misalkan:
$u = sin\ 2x$
$\dfrac{du}{dx} = 2cos\ 2x$
$\dfrac12du = cos\ 2x\ dx$
$\displaystyle \int sin^5\ 2xcos\ 2x\ dx = \displaystyle \int u^5.\dfrac12du=$
$= \displaystyle \dfrac12\int u^5du$
$= \dfrac12.\dfrac16.u^6 + C$
$= \dfrac{1}{12}u^6 + C$
$= \dfrac{1}{12}sin^6\ 2x + C$
jawab: D.
$u = sin\ 2x$
$\dfrac{du}{dx} = 2cos\ 2x$
$\dfrac12du = cos\ 2x\ dx$
$\displaystyle \int sin^5\ 2xcos\ 2x\ dx = \displaystyle \int u^5.\dfrac12du=$
$= \displaystyle \dfrac12\int u^5du$
$= \dfrac12.\dfrac16.u^6 + C$
$= \dfrac{1}{12}u^6 + C$
$= \dfrac{1}{12}sin^6\ 2x + C$
jawab: D.
14. Soal Integral UN MtkIPA 2016
Hasil dari $\displaystyle \int \dfrac{x^2 - 2}{\sqrt{6x - x^3}}\ dx = \cdots$
$A.\ -\dfrac32\sqrt{6x - x^3} + C$
$B.\ -\dfrac23\sqrt{6x - x^3} + C$
$C.\ -\dfrac16\sqrt{6x - x^3} + C$
$D.\ \dfrac16\sqrt{6x - x^3} + C$
$E.\ \dfrac23\sqrt{6x - x^3} + C$
$A.\ -\dfrac32\sqrt{6x - x^3} + C$
$B.\ -\dfrac23\sqrt{6x - x^3} + C$
$C.\ -\dfrac16\sqrt{6x - x^3} + C$
$D.\ \dfrac16\sqrt{6x - x^3} + C$
$E.\ \dfrac23\sqrt{6x - x^3} + C$
Misalkan:
$u = 6x - x^3$
$\dfrac{du}{dx} = 6 - 3x^2$
$\dfrac{du}{dx} = -3(x^2 - 2)$
$-\dfrac13du = (x^2 - 2)dx$
$\displaystyle \int \dfrac{x^2 - 2}{\sqrt{6x - x^3}}\ dx = \displaystyle \int \dfrac{1}{\sqrt{u}}.\left(-\dfrac13du\right)$
$= \displaystyle -\dfrac13\int u^{-1/2}du$
$= -\dfrac13.2.u^{1/2} + C$
$= -\dfrac23\sqrt{u} + C$
$= -\dfrac23\sqrt{6x - x^3} + C$
jawab: B.
$u = 6x - x^3$
$\dfrac{du}{dx} = 6 - 3x^2$
$\dfrac{du}{dx} = -3(x^2 - 2)$
$-\dfrac13du = (x^2 - 2)dx$
$\displaystyle \int \dfrac{x^2 - 2}{\sqrt{6x - x^3}}\ dx = \displaystyle \int \dfrac{1}{\sqrt{u}}.\left(-\dfrac13du\right)$
$= \displaystyle -\dfrac13\int u^{-1/2}du$
$= -\dfrac13.2.u^{1/2} + C$
$= -\dfrac23\sqrt{u} + C$
$= -\dfrac23\sqrt{6x - x^3} + C$
jawab: B.
15. Soal Integral UTBK MS 2019
Diketahui fungsi $f(x)$ adalah fungsi genap, jika nilai $\displaystyle \int_{-5}^5(f(x) + 3x^2)dx = 260$ dan $\displaystyle \int_2^4 f(x)dx = 2$ maka nilai $\displaystyle \int_0^2 f(x)dx + \displaystyle \int_4^5 f(x)dx = \cdots$
$A.\ -7$
$B.\ -3$
$C.\ 0$
$D.\ 3$
$E.\ 7$
$A.\ -7$
$B.\ -3$
$C.\ 0$
$D.\ 3$
$E.\ 7$
Dasar Maretong:
$\bullet$ Fungsi genap adalah fungsi yang simetris terhadap sumbu Y dimana $f(x) = f(-x)$, contohnya $y = ax^2$. Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$. Dalam hal ini $y = f(x)$ dan $y = 3x^2$ adalah fungsi genap.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
Dengan demikian:
$\displaystyle \int_{-5}^5(f(x) + 3x^2)dx = 260$
$\displaystyle 2\int_0^5(f(x) + 3x^2)dx = 260$
$\displaystyle \int_0^5(f(x) + 3x^2)dx = 130$
$\displaystyle \int_0^5 f(x)dx + \displaystyle \int_0^5 3x^2dx = 130$
$\displaystyle \int_0^5 f(x)dx + x^3\Bigr|_0^5 = 130$
$\displaystyle \int_0^5 f(x)dx + 5^3 - 0^3 = 130$
$\displaystyle \int_0^5 f(x)dx + 125 = 130$
$\displaystyle \int_0^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + \displaystyle \int_2^4 f(x)dx + \displaystyle \int_4^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + 2 + \displaystyle \int_4^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + \displaystyle \int_4^5 f(x)dx = 3$
jawab: D.
$\bullet$ Fungsi genap adalah fungsi yang simetris terhadap sumbu Y dimana $f(x) = f(-x)$, contohnya $y = ax^2$. Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$. Dalam hal ini $y = f(x)$ dan $y = 3x^2$ adalah fungsi genap.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
Dengan demikian:
$\displaystyle \int_{-5}^5(f(x) + 3x^2)dx = 260$
$\displaystyle 2\int_0^5(f(x) + 3x^2)dx = 260$
$\displaystyle \int_0^5(f(x) + 3x^2)dx = 130$
$\displaystyle \int_0^5 f(x)dx + \displaystyle \int_0^5 3x^2dx = 130$
$\displaystyle \int_0^5 f(x)dx + x^3\Bigr|_0^5 = 130$
$\displaystyle \int_0^5 f(x)dx + 5^3 - 0^3 = 130$
$\displaystyle \int_0^5 f(x)dx + 125 = 130$
$\displaystyle \int_0^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + \displaystyle \int_2^4 f(x)dx + \displaystyle \int_4^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + 2 + \displaystyle \int_4^5 f(x)dx = 5$
$\displaystyle \int_0^2 f(x)dx + \displaystyle \int_4^5 f(x)dx = 3$
jawab: D.
16. Soal Integral UTBK MS 2019
Jika nilai $\displaystyle \int_b^a f(x)dx = 5$ dan $\displaystyle \int_c^a f(x)dx = 0$, maka $\displaystyle \int_c^b f(x)dx = \cdots$
$A.\ -5$
$B.\ -3$
$C.\ 0$
$D.\ 4$
$E.\ 6$
$A.\ -5$
$B.\ -3$
$C.\ 0$
$D.\ 4$
$E.\ 6$
Dasar Maretong:
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
$\bullet$ $\displaystyle \int_a^bf(x)dx = \displaystyle -\int_b^af(x)dx$
Dengan demikian:
$\displaystyle \int_b^a f(x)dx = 5 → \displaystyle \int_a^b f(x)dx = -5$
$\displaystyle \int_c^a f(x)dx = 0 → \displaystyle \int_a^c f(x)dx = 0$
$\displaystyle \int_a^c f(x)dx = \displaystyle \int_a^b f(x)dx + \displaystyle \int_b^c f(x)dx$
$0 = -5 + \displaystyle \int_b^c f(x)dx$
$\displaystyle \int_b^c f(x)dx = 5$
$\displaystyle \int_c^b f(x)dx = -5$
jawab: A.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
$\bullet$ $\displaystyle \int_a^bf(x)dx = \displaystyle -\int_b^af(x)dx$
Dengan demikian:
$\displaystyle \int_b^a f(x)dx = 5 → \displaystyle \int_a^b f(x)dx = -5$
$\displaystyle \int_c^a f(x)dx = 0 → \displaystyle \int_a^c f(x)dx = 0$
$\displaystyle \int_a^c f(x)dx = \displaystyle \int_a^b f(x)dx + \displaystyle \int_b^c f(x)dx$
$0 = -5 + \displaystyle \int_b^c f(x)dx$
$\displaystyle \int_b^c f(x)dx = 5$
$\displaystyle \int_c^b f(x)dx = -5$
jawab: A.
17. Soal Integral UTBK MS 2019
Fungsi $f(x)$ memenuhi $f(x) = f(-x)$. Jika nilai $\displaystyle \int_{-3}^3 f(x)dx = 6$, $\displaystyle \int_{2}^3 f(x)dx = 1$, maka nilai $\displaystyle \int_{0}^2 f(x)dx = \cdots$
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
Dasar Maretong:
$\bullet$ Jika $f(x) = f(-x)$ maka fungsi tersebut adalah fungsi genap dan simetris terhadap sumbu Y. Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
Dengan demikian:
$\displaystyle \int_{-3}^3 f(x)dx = 6$
$\displaystyle 2\int_{0}^3 f(x)dx = 6$
$\displaystyle \int_{0}^3 f(x)dx = 3$
$\displaystyle \int_{0}^3 f(x)dx = \displaystyle \int_{0}^2 f(x)dx + \displaystyle \int_{2}^3 f(x)dx$
$3 = \displaystyle \int_{0}^2 f(x)dx + 1$
$\displaystyle \int_{0}^2 f(x)dx = 2$
jawab: B.
$\bullet$ Jika $f(x) = f(-x)$ maka fungsi tersebut adalah fungsi genap dan simetris terhadap sumbu Y. Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
Dengan demikian:
$\displaystyle \int_{-3}^3 f(x)dx = 6$
$\displaystyle 2\int_{0}^3 f(x)dx = 6$
$\displaystyle \int_{0}^3 f(x)dx = 3$
$\displaystyle \int_{0}^3 f(x)dx = \displaystyle \int_{0}^2 f(x)dx + \displaystyle \int_{2}^3 f(x)dx$
$3 = \displaystyle \int_{0}^2 f(x)dx + 1$
$\displaystyle \int_{0}^2 f(x)dx = 2$
jawab: B.
18. Soal Integral UTBK MS 2019
Misalkan fungsi $f$ memenuhi $f(x + 5) = f(x)$ untuk setiap $x \in R$. Jika $\displaystyle \int_1^5 f(x)dx = 3$ dan $\displaystyle \int_{-5}^{-4} f(x)dx = -2$ maka nilai $\displaystyle \int_5^{15} f(x)dx = \cdots$
Dasar Maretong:
Jika $f(x) = f(x + c)$ maka $f(x)$ adalah fungsi periodik dengan periode $c$, sehingga berlaku:
$\displaystyle \int_a^b f(x)dx = \displaystyle \int_{a + c}^{b + c}f(x)dx = \displaystyle \int_{a + 2c}^{b + 2c}f(x)dx = \cdots$
$f(x) = f(x + 5)$, Berarti $f(x)$ adalah fungsi periodik dengan periode 5.
Dengan demikian:
$\displaystyle \int_1^5 f(x)dx$ $= \displaystyle \int_6^{10} f(x)dx$ $= \displaystyle \int_{11}^{15} f(x)dx = 3$
$\displaystyle \int_{-5}^{-4} f(x)dx$ $= \displaystyle \int_0^{1} f(x)dx$ $= \displaystyle \int_5^6 f(x)dx$ $= \displaystyle \int_{10}^{11} f(x)dx = -2$
$\displaystyle \int_5^{15} f(x)dx = \displaystyle \int_5^6 f(x)dx$ $+ \displaystyle \int_6^{10} f(x)dx + \displaystyle \int_{10}^{11} f(x)dx + \displaystyle \int_{11}^{15} f(x)dx$
$= -2 + 3 -2 + 3$
$= 2$
jawab: D.
Jika $f(x) = f(x + c)$ maka $f(x)$ adalah fungsi periodik dengan periode $c$, sehingga berlaku:
$\displaystyle \int_a^b f(x)dx = \displaystyle \int_{a + c}^{b + c}f(x)dx = \displaystyle \int_{a + 2c}^{b + 2c}f(x)dx = \cdots$
$f(x) = f(x + 5)$, Berarti $f(x)$ adalah fungsi periodik dengan periode 5.
Dengan demikian:
$\displaystyle \int_1^5 f(x)dx$ $= \displaystyle \int_6^{10} f(x)dx$ $= \displaystyle \int_{11}^{15} f(x)dx = 3$
$\displaystyle \int_{-5}^{-4} f(x)dx$ $= \displaystyle \int_0^{1} f(x)dx$ $= \displaystyle \int_5^6 f(x)dx$ $= \displaystyle \int_{10}^{11} f(x)dx = -2$
$\displaystyle \int_5^{15} f(x)dx = \displaystyle \int_5^6 f(x)dx$ $+ \displaystyle \int_6^{10} f(x)dx + \displaystyle \int_{10}^{11} f(x)dx + \displaystyle \int_{11}^{15} f(x)dx$
$= -2 + 3 -2 + 3$
$= 2$
jawab: D.
19. Soal Integral UTBK MS 2019
Diketahui $f(x)$ merupakan fungsi genap, jika $\displaystyle \int_{-4}^{4} f(x)dx = 16$, $\displaystyle \int_{3}^{4} f(2x - 2)dx = 11$ dan $\displaystyle \int_{-5}^{-1} f(1 - x)dx = 6$, maka $\displaystyle \int_{0}^{2} f(x)dx = \cdots$
A. 22
B. 23
C. 24
D. 25
E. 26
A. 22
B. 23
C. 24
D. 25
E. 26
Dasar Maretong:
$\bullet$ Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
$\displaystyle \int_{-4}^{4} f(x)dx = 16$
$\displaystyle 2\int_{0}^{4} f(x)dx = 16$
$\displaystyle \int_{0}^{4} f(x)dx = 8$
$\displaystyle \int_{3}^{4} f(2x - 2)dx = 11$
Misalkan:
$u = 2x - 2$
$\dfrac{du}{dx} = 2$
$\dfrac12du = dx$
$x = 3 → u = 4;\ x = 4 → u = 6$
$\displaystyle \int_{3}^{4} f(2x - 2)dx = 11$
$\displaystyle \dfrac12\int_{4}^{6} f(u)du = 11$
$\displaystyle \int_{4}^{6} f(u)du = 22$
$\displaystyle \int_{4}^{6} f(x)dx = 22$
$\displaystyle \int_{-5}^{-1} f(1 - x)dx = 6$
Misalkan:
$u = 1 - x$
$-du = dx$
$x = -1 → u = 2;\ x = -5 → u = 6$
$\displaystyle \int_{-5}^{-1} f(1 - x)dx = 6$
$\displaystyle -\int_{6}^{2} f(u)du = 6$
$\displaystyle \int_{2}^{6} f(u)du = 6$
$\displaystyle \int_{2}^{6} f(x)dx = 6$
$\displaystyle \int_{2}^{6} f(x)dx = \displaystyle \int_{2}^{4} f(x)dx + \displaystyle \int_{4}^{6} f(x)dx$
$6 = \displaystyle \int_{2}^{4} f(x)dx + 22$
$\displaystyle \int_{2}^{4} f(x)dx = -16$
$\displaystyle \int_{0}^{4} f(x)dx = \displaystyle \int_{0}^{2} f(x)dx + \displaystyle \int_{2}^{4} f(x)dx$
$8 = \displaystyle \int_{0}^{2} f(x)dx - 16$
$\displaystyle \int_{0}^{2} f(x)dx = 24$
jawab: C.
$\bullet$ Jika $f(x)$ adalah fungsi genap maka $\displaystyle \int_{-a}^a f(x) dx = \displaystyle 2\int_0^a f(x)dx$.
$\bullet$ $\displaystyle \int_a^cf(x)dx = \displaystyle \int_a^bf(x)dx + \displaystyle \int_b^cf(x)dx$
$\displaystyle \int_{-4}^{4} f(x)dx = 16$
$\displaystyle 2\int_{0}^{4} f(x)dx = 16$
$\displaystyle \int_{0}^{4} f(x)dx = 8$
$\displaystyle \int_{3}^{4} f(2x - 2)dx = 11$
Misalkan:
$u = 2x - 2$
$\dfrac{du}{dx} = 2$
$\dfrac12du = dx$
$x = 3 → u = 4;\ x = 4 → u = 6$
$\displaystyle \int_{3}^{4} f(2x - 2)dx = 11$
$\displaystyle \dfrac12\int_{4}^{6} f(u)du = 11$
$\displaystyle \int_{4}^{6} f(u)du = 22$
$\displaystyle \int_{4}^{6} f(x)dx = 22$
$\displaystyle \int_{-5}^{-1} f(1 - x)dx = 6$
Misalkan:
$u = 1 - x$
$-du = dx$
$x = -1 → u = 2;\ x = -5 → u = 6$
$\displaystyle \int_{-5}^{-1} f(1 - x)dx = 6$
$\displaystyle -\int_{6}^{2} f(u)du = 6$
$\displaystyle \int_{2}^{6} f(u)du = 6$
$\displaystyle \int_{2}^{6} f(x)dx = 6$
$\displaystyle \int_{2}^{6} f(x)dx = \displaystyle \int_{2}^{4} f(x)dx + \displaystyle \int_{4}^{6} f(x)dx$
$6 = \displaystyle \int_{2}^{4} f(x)dx + 22$
$\displaystyle \int_{2}^{4} f(x)dx = -16$
$\displaystyle \int_{0}^{4} f(x)dx = \displaystyle \int_{0}^{2} f(x)dx + \displaystyle \int_{2}^{4} f(x)dx$
$8 = \displaystyle \int_{0}^{2} f(x)dx - 16$
$\displaystyle \int_{0}^{2} f(x)dx = 24$
jawab: C.
20. Soal Integral SIMAK UI MtkIPA 2019
Jika $\displaystyle \int_{a}^{b} f'(x)f(x)d(x) = 10$ dan $f(a) = 2 + f(b)$, nilai $f(b) =$ . . . .
$A.\ -2$
$B.\ -4$
$C.\ -6$
$D.\ -8$
$E.\ -10$
$A.\ -2$
$B.\ -4$
$C.\ -6$
$D.\ -8$
$E.\ -10$
$f(a) = 2 + f(b)$
$f(b) - f(a) = -2$ . . . . *
$\displaystyle \int_{a}^{b} f'(x)f(x)d(x) = 10$
$\displaystyle \int_{a}^{b} f(x)d(f(x)) = 10$
$\left.\begin{matrix}\dfrac12 f^2(x) \end{matrix}\right|_a^b = 10$
$\dfrac12(f^2(b) - f^2(a)) = 10$
$(f(b) + f(a))(f(b) - f(a)) = 20$ . . . . **
Masukkan pers * ke pers **
$(f(b) + f(a)).(-2) = 20$
$f(b) + f(a) = -10$ . . . . ***
Eliminasi pers * dan pers ***
$f(b) - f(a) = -2$
$f(b) + f(a) = -10$
$....................\ +$
$2f(b) = -12$
$f(b) = -6$
jawab: C.
$f(b) - f(a) = -2$ . . . . *
$\displaystyle \int_{a}^{b} f'(x)f(x)d(x) = 10$
$\displaystyle \int_{a}^{b} f(x)d(f(x)) = 10$
$\left.\begin{matrix}\dfrac12 f^2(x) \end{matrix}\right|_a^b = 10$
$\dfrac12(f^2(b) - f^2(a)) = 10$
$(f(b) + f(a))(f(b) - f(a)) = 20$ . . . . **
Masukkan pers * ke pers **
$(f(b) + f(a)).(-2) = 20$
$f(b) + f(a) = -10$ . . . . ***
Eliminasi pers * dan pers ***
$f(b) - f(a) = -2$
$f(b) + f(a) = -10$
$....................\ +$
$2f(b) = -12$
$f(b) = -6$
jawab: C.
21. Soal Integral SBMPTN MtkIPA 2018
Nilai $\displaystyle \int_{1/8}^{1/3} \dfrac{3}{x^2}\sqrt{1 + \dfrac 1x}\ dx$ adalah . . . .
A. 19
B. 38
C. 57
D. 76
E. 95
A. 19
B. 38
C. 57
D. 76
E. 95
Misalkan:
$u = 1 + \dfrac 1x$
$\dfrac{du}{dx} = -\dfrac{1}{x^2}$
$-du = \dfrac{1}{x^2}dx$
$x = \dfrac13 → u = 4$
$x = \dfrac18 → u = 9$
$\displaystyle \int_{1/8}^{1/3} \dfrac{3}{x^2}\sqrt{1 + \dfrac 1x}\ dx = \displaystyle -3\int_{9}^{4} \sqrt{u}\ du$
$= \displaystyle 3\int_{4}^{9} \sqrt{u}\ du$
$= 3.\dfrac23.u^{3/2}\Bigr|_4^9$
$= 2.\left[(9)^{3/2} - (4)^{3/2}\right]$
$= 2.\left[3^3 - 2^3\right]$
$= 2.(27 - 8)$
$= 2.19$
$= 38$
jawab: B.
$u = 1 + \dfrac 1x$
$\dfrac{du}{dx} = -\dfrac{1}{x^2}$
$-du = \dfrac{1}{x^2}dx$
$x = \dfrac13 → u = 4$
$x = \dfrac18 → u = 9$
$\displaystyle \int_{1/8}^{1/3} \dfrac{3}{x^2}\sqrt{1 + \dfrac 1x}\ dx = \displaystyle -3\int_{9}^{4} \sqrt{u}\ du$
$= \displaystyle 3\int_{4}^{9} \sqrt{u}\ du$
$= 3.\dfrac23.u^{3/2}\Bigr|_4^9$
$= 2.\left[(9)^{3/2} - (4)^{3/2}\right]$
$= 2.\left[3^3 - 2^3\right]$
$= 2.(27 - 8)$
$= 2.19$
$= 38$
jawab: B.
22. Soal Integral SBMPTN MtkIPA 2018
Nilai $\displaystyle \int_1^{36} \dfrac{3}{\sqrt{x}(3 + \sqrt{x})^{3/2}}\ dx$ adalah . . . .
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
Misalkan:
$u = 3 + \sqrt{x}$
$\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}$
$2du = \dfrac{1}{\sqrt{x}}dx$
$(x = 1 → u = 4;\ x = 36 → u = 9)$
$\displaystyle \int_1^{36} \dfrac{3}{\sqrt{x}(3 + \sqrt{x})^{3/2}}\ dx = \displaystyle \int_4^9 \dfrac{3}{u^{3/2}}.2du$
$= \displaystyle \int_4^9 \dfrac{6}{u^{3/2}}du$
$= \displaystyle \int_4^9 6u^{-3/2}du$
$= -12u^{-1/2}\Bigr|_4^9$
$= -12(9^{-1/2} - 4^{-1/2})$
$= -12\left(\dfrac13 - \dfrac12\right)$
$= -12.\left(-\dfrac16\right)$
$= 2$
jawab: B.
$u = 3 + \sqrt{x}$
$\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}$
$2du = \dfrac{1}{\sqrt{x}}dx$
$(x = 1 → u = 4;\ x = 36 → u = 9)$
$\displaystyle \int_1^{36} \dfrac{3}{\sqrt{x}(3 + \sqrt{x})^{3/2}}\ dx = \displaystyle \int_4^9 \dfrac{3}{u^{3/2}}.2du$
$= \displaystyle \int_4^9 \dfrac{6}{u^{3/2}}du$
$= \displaystyle \int_4^9 6u^{-3/2}du$
$= -12u^{-1/2}\Bigr|_4^9$
$= -12(9^{-1/2} - 4^{-1/2})$
$= -12\left(\dfrac13 - \dfrac12\right)$
$= -12.\left(-\dfrac16\right)$
$= 2$
jawab: B.
23. Soal Integral SBMPTN MtkIPA 2018
Nilai $\displaystyle \int_0^3 \dfrac{3x}{\sqrt{x + 1}}\ dx$ adalah . . . .
A. 3
B. 6
C. 8
D. 9
E. 12
A. 3
B. 6
C. 8
D. 9
E. 12
Misalkan:
$u = x + 1 → x = u - 1$
$\dfrac{du}{dx} = 1 → du = dx$
$x = 0 → u = 1;\ x = 3 → u = 4$
$\displaystyle \int_0^3 \dfrac{3x}{\sqrt{x + 1}}\ dx = \displaystyle \int_1^4 \dfrac{3(u - 1)}{\sqrt{u}}\ du$
$= \displaystyle \int_1^4 \dfrac{3u - 3}{\sqrt{u}}\ du$
$= \displaystyle \int_1^4 \dfrac{3u}{\sqrt{u}}du - \displaystyle \int_1^4 \dfrac{3}{\sqrt{u}}du$
$= \displaystyle \int_1^4 3u^{1/2}du - \displaystyle \int_1^4 3u^{-1/2}du$
$= 3.\dfrac23.u^{3/2}\Bigr|_1^4 - 3.2.u^{1/2}\Bigr|_1^4$
$= 2.(4^{3/2} - 1^{3/2}) - 6.(4^{1/2} - 1^{1/2})$
$= 2.(8 - 1) - 6.(2 - 1)$
$= 2.7 - 6.1$
$= 14 - 6$
$= 8$
jawab: C.
$u = x + 1 → x = u - 1$
$\dfrac{du}{dx} = 1 → du = dx$
$x = 0 → u = 1;\ x = 3 → u = 4$
$\displaystyle \int_0^3 \dfrac{3x}{\sqrt{x + 1}}\ dx = \displaystyle \int_1^4 \dfrac{3(u - 1)}{\sqrt{u}}\ du$
$= \displaystyle \int_1^4 \dfrac{3u - 3}{\sqrt{u}}\ du$
$= \displaystyle \int_1^4 \dfrac{3u}{\sqrt{u}}du - \displaystyle \int_1^4 \dfrac{3}{\sqrt{u}}du$
$= \displaystyle \int_1^4 3u^{1/2}du - \displaystyle \int_1^4 3u^{-1/2}du$
$= 3.\dfrac23.u^{3/2}\Bigr|_1^4 - 3.2.u^{1/2}\Bigr|_1^4$
$= 2.(4^{3/2} - 1^{3/2}) - 6.(4^{1/2} - 1^{1/2})$
$= 2.(8 - 1) - 6.(2 - 1)$
$= 2.7 - 6.1$
$= 14 - 6$
$= 8$
jawab: C.
24. Soal Integral SBMPTN MtkIPA 2018
Jika $\displaystyle \int_1^2 f(x)dx = \sqrt{2}$, maka nilai $\displaystyle \int_1^4 \dfrac{1}{\sqrt{x}}f(\sqrt{x})dx$ adalah . . . .
$A.\ \dfrac{\sqrt{2}}{4}$
$B.\ \dfrac{\sqrt{2}}{2}$
$C.\ \sqrt{2}$
$D.\ 2\sqrt{2}$
$E.\ 4\sqrt{2}$
$A.\ \dfrac{\sqrt{2}}{4}$
$B.\ \dfrac{\sqrt{2}}{2}$
$C.\ \sqrt{2}$
$D.\ 2\sqrt{2}$
$E.\ 4\sqrt{2}$
Dasar Maretong:
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = \sqrt{x}$
$\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}$
$2du = \dfrac{1}{\sqrt{x}}dx$
$x = 1 → u = 1;\ x = 4 → u = 2$
$\displaystyle \int_1^4 \dfrac{1}{\sqrt{x}}f(\sqrt{x})dx = \displaystyle \int_1^2 f(u).2du$
$= 2\displaystyle \int_1^2 f(u)du$
$= 2\displaystyle \int_1^2 f(x)dx$
$= 2.\sqrt{2}$
$= 2\sqrt{2}$
jawab: D.
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = \sqrt{x}$
$\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}$
$2du = \dfrac{1}{\sqrt{x}}dx$
$x = 1 → u = 1;\ x = 4 → u = 2$
$\displaystyle \int_1^4 \dfrac{1}{\sqrt{x}}f(\sqrt{x})dx = \displaystyle \int_1^2 f(u).2du$
$= 2\displaystyle \int_1^2 f(u)du$
$= 2\displaystyle \int_1^2 f(x)dx$
$= 2.\sqrt{2}$
$= 2\sqrt{2}$
jawab: D.
25. Soal Integral SBMPTN MtkIPA 2018
Nilai $\displaystyle \int_0^2 (3x + 9)\sqrt{x^2 + 6x}\ dx$ adalah . . . .
A. 4
B. 8
C. 16
D. 32
E. 64
A. 4
B. 8
C. 16
D. 32
E. 64
Misalkan:
$u = x^2 + 6x$
$\dfrac{du}{dx} = 2x + 6$
$du = (2x + 6)dx$
$\dfrac32du = (3x + 9)dx$
$x = 0 → u = 0;\ x = 2 → u = 16$
$\displaystyle \int_0^2 (3x + 9)\sqrt{x^2 + 6x}\ dx = \displaystyle \int_0^{16}\sqrt{u}.\dfrac32du$
$= \displaystyle \dfrac32\int_0^{16}\sqrt{u}\ du$
$= \dfrac32.\dfrac23.u^{3/2}\Bigr|_0^{16}$
$= u^{3/2}\Bigr|_0^{16}$
$= 16^{3/2} - 0^{3/2}$
$= 4^3 - 0$
$= 64$
jawab: E.
$u = x^2 + 6x$
$\dfrac{du}{dx} = 2x + 6$
$du = (2x + 6)dx$
$\dfrac32du = (3x + 9)dx$
$x = 0 → u = 0;\ x = 2 → u = 16$
$\displaystyle \int_0^2 (3x + 9)\sqrt{x^2 + 6x}\ dx = \displaystyle \int_0^{16}\sqrt{u}.\dfrac32du$
$= \displaystyle \dfrac32\int_0^{16}\sqrt{u}\ du$
$= \dfrac32.\dfrac23.u^{3/2}\Bigr|_0^{16}$
$= u^{3/2}\Bigr|_0^{16}$
$= 16^{3/2} - 0^{3/2}$
$= 4^3 - 0$
$= 64$
jawab: E.
26. Soal Integral SBMPTN MtkIPA 2018
Jika $\displaystyle \int_0^4 f(x)dx = \sqrt{2}$ maka nilai $\displaystyle \int_0^2 xf(x^2)dx$ adalah . . . .
$A.\ \dfrac{\sqrt{2}}{4}$
$B.\ \dfrac{\sqrt{2}}{2}$
$C.\ \sqrt{2}$
$D.\ 2\sqrt{2}$
$E.\ 4\sqrt{2}$
$A.\ \dfrac{\sqrt{2}}{4}$
$B.\ \dfrac{\sqrt{2}}{2}$
$C.\ \sqrt{2}$
$D.\ 2\sqrt{2}$
$E.\ 4\sqrt{2}$
Dasar Maretong:
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = x^2$
$\dfrac{du}{dx} = 2x$
$\dfrac12du = xdx$
$x = 0 → u = 0;\ x = 2 → x = 4$
$\displaystyle \int_0^2 xf(x^2)dx = \displaystyle \int_0^4 f(u).\dfrac12du$
$= \displaystyle \dfrac12\int_0^4 f(u)du$
$= \displaystyle \dfrac12\int_0^4 f(x)dx$
$= \dfrac12.\sqrt{2}$
$= \dfrac{\sqrt{2}}{2}$
jawab: B.
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = x^2$
$\dfrac{du}{dx} = 2x$
$\dfrac12du = xdx$
$x = 0 → u = 0;\ x = 2 → x = 4$
$\displaystyle \int_0^2 xf(x^2)dx = \displaystyle \int_0^4 f(u).\dfrac12du$
$= \displaystyle \dfrac12\int_0^4 f(u)du$
$= \displaystyle \dfrac12\int_0^4 f(x)dx$
$= \dfrac12.\sqrt{2}$
$= \dfrac{\sqrt{2}}{2}$
jawab: B.
27. Soal Integral Utul MtkIPA 2016
$\displaystyle \int_{1/2}^1 (\sqrt[3]{2x - 1} + sin\ \pi x)dx = \dots$
$A.\ \dfrac{3\pi - 8}{8\pi}$
$B.\ \dfrac{3\pi - 4}{4\pi}$
$C.\ \dfrac{3\pi + 4}{4\pi}$
$D.\ \dfrac{3\pi + 8}{8\pi}$
$E.\ \dfrac34 + \pi$
$A.\ \dfrac{3\pi - 8}{8\pi}$
$B.\ \dfrac{3\pi - 4}{4\pi}$
$C.\ \dfrac{3\pi + 4}{4\pi}$
$D.\ \dfrac{3\pi + 8}{8\pi}$
$E.\ \dfrac34 + \pi$
Dasar Maretong:
$\displaystyle \int m(ax + b)^n\ dx = \dfrac{m}{a(n + 1)}(ax + b)^{n + 1}$
$\displaystyle \int_{1/2}^1 (\sqrt[3]{2x - 1} + sin\ \pi x)dx = \displaystyle \int_{1/2}^1 ((2x - 1)^{1/3} + sin\ \pi x)dx$
$= \left[\dfrac{1}{2\left(1 + \dfrac13\right)}.(2x - 1)^{1 + 1/3} - \dfrac{1}{\pi}.cos\ \pi x\right]_{1/2}^1$
$= \left(\dfrac38(2x - 1)^{4/3} - \dfrac{1}{\pi}cos\ \pi x\right)\Bigr|_{1/2}^1$
$= \dfrac38.(2.1 - 1)^{4/3} - \dfrac{1}{\pi}cos\ \pi.1 - \left(\left(2.\dfrac12 - 1\right)^{4/3} - \dfrac{1}{\pi}cos\ \dfrac{\pi}{2}\right)$
$= \dfrac38.1^{4/3} - \dfrac{1}{\pi}.(-1) - \left(0^{4/3} - \dfrac{1}{\pi}.0\right)$
$= \dfrac38.1 + \dfrac{1}{\pi}$
$= \dfrac38 + \dfrac{1}{\pi}$
$= \dfrac{3\pi + 8}{8\pi}$
jawab: D.
$\displaystyle \int m(ax + b)^n\ dx = \dfrac{m}{a(n + 1)}(ax + b)^{n + 1}$
$\displaystyle \int_{1/2}^1 (\sqrt[3]{2x - 1} + sin\ \pi x)dx = \displaystyle \int_{1/2}^1 ((2x - 1)^{1/3} + sin\ \pi x)dx$
$= \left[\dfrac{1}{2\left(1 + \dfrac13\right)}.(2x - 1)^{1 + 1/3} - \dfrac{1}{\pi}.cos\ \pi x\right]_{1/2}^1$
$= \left(\dfrac38(2x - 1)^{4/3} - \dfrac{1}{\pi}cos\ \pi x\right)\Bigr|_{1/2}^1$
$= \dfrac38.(2.1 - 1)^{4/3} - \dfrac{1}{\pi}cos\ \pi.1 - \left(\left(2.\dfrac12 - 1\right)^{4/3} - \dfrac{1}{\pi}cos\ \dfrac{\pi}{2}\right)$
$= \dfrac38.1^{4/3} - \dfrac{1}{\pi}.(-1) - \left(0^{4/3} - \dfrac{1}{\pi}.0\right)$
$= \dfrac38.1 + \dfrac{1}{\pi}$
$= \dfrac38 + \dfrac{1}{\pi}$
$= \dfrac{3\pi + 8}{8\pi}$
jawab: D.
28. Soal Integral SIMAK UI MtkIPA 2018
Jika $f(x)$ fungsi kontinu di interval [1, 30] dan $\displaystyle \int_6^{30} f(x)dx = 30$, maka $\displaystyle \int_1^9 f(3y + 3)dy = \cdots$
A. 5
B. 10
C. 15
D. 18
E. 27
A. 5
B. 10
C. 15
D. 18
E. 27
Dasar Maretong:
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = 3y + 3$
$\dfrac{du}{dy} = 3$
$\dfrac13du = dy$
$y = 1 → u = 6;\ y = 9 → u = 30$
$\displaystyle \int_1^9 f(3y + 3)dy = \displaystyle \int_6^{30} f(u).\dfrac13du$
$= \dfrac13\displaystyle \int_6^{30} f(u)du$
$= \dfrac13\displaystyle \int_6^{30} f(x)dx$
$= \dfrac13.30$
$= 10$
jawab: B.
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
Misalkan:
$u = 3y + 3$
$\dfrac{du}{dy} = 3$
$\dfrac13du = dy$
$y = 1 → u = 6;\ y = 9 → u = 30$
$\displaystyle \int_1^9 f(3y + 3)dy = \displaystyle \int_6^{30} f(u).\dfrac13du$
$= \dfrac13\displaystyle \int_6^{30} f(u)du$
$= \dfrac13\displaystyle \int_6^{30} f(x)dx$
$= \dfrac13.30$
$= 10$
jawab: B.
29. Soal Integral SIMAK UI MtkIPA 2017
Jika $3x^5 - 3 = \displaystyle \int_c^x g(t)dt$, maka $g\left(\dfrac c2\right) = \cdots$
$A.\ \dfrac{10}{16}$
$B.\ \dfrac{12}{16}$
$C.\ \dfrac{14}{16}$
$D.\ \dfrac{15}{16}$
$E.\ \dfrac{17}{16}$
$A.\ \dfrac{10}{16}$
$B.\ \dfrac{12}{16}$
$C.\ \dfrac{14}{16}$
$D.\ \dfrac{15}{16}$
$E.\ \dfrac{17}{16}$
Dasar Maretong:
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
$G'(x) = g(x)$
$3x^5 - 3 = \displaystyle \int_c^x g(t)dt$
$3x^5 - 3 = G(x)\Bigr|_c^x$
$3x^5 - 3 = G(x) - G(c)$
$15x^4 = G'(x)$
$15x^4 = g(x)$
$3x^5 - 3 = \displaystyle \int_c^x g(t)dt$
$3x^5 - 3 = \displaystyle \int_c^x g(x)dx$
$3x^5 - 3 = \displaystyle \int_c^x 15x^4\ dx$
$3x^5 - 3 = 3x^5\Bigr|_c^x$
$3x^5 - 3 = 3x^5 - 3c^5$
Kesamaan:
$-3 = -3c^5$
$1 = c^5$
$c = 1$
$g(x) = 15x^4$
$g\left(\dfrac c2\right) = g\left(\dfrac12\right) = 15.\left(\dfrac12\right)^4 = \dfrac{15}{16}$
jawab: D.
$\displaystyle \int_p^q f(x)dx = \displaystyle \int_p^q f(u)du = \displaystyle \int_p^q f(t)dt = \cdots$
$G'(x) = g(x)$
$3x^5 - 3 = \displaystyle \int_c^x g(t)dt$
$3x^5 - 3 = G(x)\Bigr|_c^x$
$3x^5 - 3 = G(x) - G(c)$
$15x^4 = G'(x)$
$15x^4 = g(x)$
$3x^5 - 3 = \displaystyle \int_c^x g(t)dt$
$3x^5 - 3 = \displaystyle \int_c^x g(x)dx$
$3x^5 - 3 = \displaystyle \int_c^x 15x^4\ dx$
$3x^5 - 3 = 3x^5\Bigr|_c^x$
$3x^5 - 3 = 3x^5 - 3c^5$
Kesamaan:
$-3 = -3c^5$
$1 = c^5$
$c = 1$
$g(x) = 15x^4$
$g\left(\dfrac c2\right) = g\left(\dfrac12\right) = 15.\left(\dfrac12\right)^4 = \dfrac{15}{16}$
jawab: D.
30. Soal Integral SBMPTN MtkIPA 2017
Jika $\displaystyle \int_{-4}^{4} f(x)(sin\ x + 1)dx = 8$, dengan $f(x)$ fungsi genap dan $\displaystyle \int_{-2}^{4} f(x)dx = 4$ maka $\displaystyle \int_{-2}{0} f(x)dx = \cdots$
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
Dasar Maretong:
$\bullet$ Fungsi ganjil adalah fungsi yang simetris terhadap titik O(0, 0) dan berlaku $f(-x) = -f(x)$, contohnya $y = sin\ x$ dan $y = x$.
$\bullet$ Untuk fungsi ganjil berlaku: $\displaystyle \int_{-a}^{a}f(x)dx = 0$
$\displaystyle \int_{-4}^{4} f(x)sin\ x\ dx + \displaystyle \int_{-4}^{4} f(x)dx = 8$
Karena $sin\ x$ merupakan fungsi ganjil, maka $f(x).sin\ x$ merupakan fungsi ganjil, sehingga $\displaystyle \int_{-4}^{4} f(x)sin\ x\ dx = 0$
Dengan demikian:
$\displaystyle \int_{-4}^{4} f(x)dx = 8$
$\displaystyle 2\int_{0}^{4} f(x)dx = 8$
$\displaystyle \int_{0}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{0} f(x)dx + \displaystyle \int_{0}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{0} f(x)dx + 4 = 4$
$\displaystyle \int_{-2}^{0} f(x)dx = 0$
jawab: A.
$\bullet$ Fungsi ganjil adalah fungsi yang simetris terhadap titik O(0, 0) dan berlaku $f(-x) = -f(x)$, contohnya $y = sin\ x$ dan $y = x$.
$\bullet$ Untuk fungsi ganjil berlaku: $\displaystyle \int_{-a}^{a}f(x)dx = 0$
$\displaystyle \int_{-4}^{4} f(x)sin\ x\ dx + \displaystyle \int_{-4}^{4} f(x)dx = 8$
Karena $sin\ x$ merupakan fungsi ganjil, maka $f(x).sin\ x$ merupakan fungsi ganjil, sehingga $\displaystyle \int_{-4}^{4} f(x)sin\ x\ dx = 0$
Dengan demikian:
$\displaystyle \int_{-4}^{4} f(x)dx = 8$
$\displaystyle 2\int_{0}^{4} f(x)dx = 8$
$\displaystyle \int_{0}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{0} f(x)dx + \displaystyle \int_{0}^{4} f(x)dx = 4$
$\displaystyle \int_{-2}^{0} f(x)dx + 4 = 4$
$\displaystyle \int_{-2}^{0} f(x)dx = 0$
jawab: A.
31. Soal Integral SBMPTN MtkIPA 2016
Diketahui fungsi $f(x) = f(x + 2)$ untuk setiap $x$. Jika $\displaystyle \int_0^2 f(x)dx = B$, maka $\displaystyle \int_3^7 f(x + 8)dx = \cdots$
A. B
B. 2B
C. 3B
D. 4B
E. 5B
A. B
B. 2B
C. 3B
D. 4B
E. 5B
Dasar Maretong:
Jika $f(x) = f(x + c)$ maka $f(x)$ adalah fungsi periodik dengan periode $c$, sehingga berlaku:
$\bullet$ $\displaystyle \int_a^b f(x)dx = \displaystyle \int_{a + c}^{b + c}f(x)dx = \displaystyle \int_{a + 2c}^{b + 2c}f(x)dx = \cdots$
$\bullet$ $\displaystyle \int_a^b f(x)dx = \displaystyle \int_a^b f(x + c)dx = \displaystyle \int_a^b f(x + 2c)dx = \cdots$
$\displaystyle \int_3^7 f(x + 8)dx = \displaystyle \int_3^7 f(x)dx$
$= \displaystyle \int_3^4 f(x)dx + \displaystyle \int_4^6 f(x)dx + \displaystyle \int_6^7 f(x)dx$
$= \displaystyle \int_{3 - 2}^{4 - 2} f(x)dx + \displaystyle \int_{4 - 2.2}^{6 - 2.2} f(x)dx + \displaystyle \int_{6 - 3.2}^{7 - 3.2} f(x)dx$
$= \displaystyle \int_1^2 f(x)dx + \displaystyle \int_0^2 f(x)dx + \displaystyle \int_0^1 f(x)dx$
$= \displaystyle \int_0^1 f(x)dx + \displaystyle \int_1^2 f(x)dx + \displaystyle \int_0^2 f(x)dx$
$= \displaystyle \int_0^2 f(x)dx + \displaystyle \int_0^2 f(x)dx$
$= B + B$
$= 2B$
jawab: B.
Jika $f(x) = f(x + c)$ maka $f(x)$ adalah fungsi periodik dengan periode $c$, sehingga berlaku:
$\bullet$ $\displaystyle \int_a^b f(x)dx = \displaystyle \int_{a + c}^{b + c}f(x)dx = \displaystyle \int_{a + 2c}^{b + 2c}f(x)dx = \cdots$
$\bullet$ $\displaystyle \int_a^b f(x)dx = \displaystyle \int_a^b f(x + c)dx = \displaystyle \int_a^b f(x + 2c)dx = \cdots$
$\displaystyle \int_3^7 f(x + 8)dx = \displaystyle \int_3^7 f(x)dx$
$= \displaystyle \int_3^4 f(x)dx + \displaystyle \int_4^6 f(x)dx + \displaystyle \int_6^7 f(x)dx$
$= \displaystyle \int_{3 - 2}^{4 - 2} f(x)dx + \displaystyle \int_{4 - 2.2}^{6 - 2.2} f(x)dx + \displaystyle \int_{6 - 3.2}^{7 - 3.2} f(x)dx$
$= \displaystyle \int_1^2 f(x)dx + \displaystyle \int_0^2 f(x)dx + \displaystyle \int_0^1 f(x)dx$
$= \displaystyle \int_0^1 f(x)dx + \displaystyle \int_1^2 f(x)dx + \displaystyle \int_0^2 f(x)dx$
$= \displaystyle \int_0^2 f(x)dx + \displaystyle \int_0^2 f(x)dx$
$= B + B$
$= 2B$
jawab: B.
32. Soal Integral SBMPTN MtkIPA 2016
Nilai $k$ antara $0\ dan\ \pi$ yang membuat $\displaystyle \int_0^k (sin\ x + cos\ x)dx$ maksimum adalah . . . .
$A.\ \dfrac{5\pi}{4}$
$B.\ \dfrac{3\pi}{4}$
$C.\ \dfrac{2\pi}{3}$
$D.\ \dfrac{\pi}{3}$
$E.\ \dfrac{\pi}{2}$
$A.\ \dfrac{5\pi}{4}$
$B.\ \dfrac{3\pi}{4}$
$C.\ \dfrac{2\pi}{3}$
$D.\ \dfrac{\pi}{3}$
$E.\ \dfrac{\pi}{2}$
$\displaystyle \int_0^k (sin\ x + cos\ x)dx$
$= (-cos\ x + sin\ x)\Bigr|_0^k$
$= -cos\ k + sin\ k - (-cos\ 0 + sin\ 0)$
$= sin\ k - cos\ k - (-1 + 0)$
$= sin\ k - cos\ k + 1$
$f(k) = sin\ k - cos\ k + 1$
$f'(k) = 0$
$sin\ k + cos\ k = 0$
$sin\ k = -cos\ k$
$\dfrac{sin\ k}{cos\ k} = -1$
$tan\ k = -1$
$k = \dfrac{3\pi}{4}$
jawab: B.
$= (-cos\ x + sin\ x)\Bigr|_0^k$
$= -cos\ k + sin\ k - (-cos\ 0 + sin\ 0)$
$= sin\ k - cos\ k - (-1 + 0)$
$= sin\ k - cos\ k + 1$
$f(k) = sin\ k - cos\ k + 1$
$f'(k) = 0$
$sin\ k + cos\ k = 0$
$sin\ k = -cos\ k$
$\dfrac{sin\ k}{cos\ k} = -1$
$tan\ k = -1$
$k = \dfrac{3\pi}{4}$
jawab: B.
33. Soal Integral SBMPTN MtkIPA 2016
Diketahui $f(x) = k(x^3 - 6x^2 + 9x),\ k > 0$ dan $\displaystyle \int_0^a f(x)dx = 27$ untuk $(a,\ b)$ titik balik minimum. Nilai $k$ adalah . . . .
A. 9
B. 8
C. 6
D. 4
E. 3
A. 9
B. 8
C. 6
D. 4
E. 3
$f(x) = k(x^3 - 6x^2 + 9x)$
$f'(x) = 0$
$k(3x^2 - 12x + 9) = 0$
$3k(x^2 - 4x + 3) = 0$
$x^2 - 4x + 3 = 0$
$(x - 3)(x - 1) = 0$
$x = 3\ atau\ x = 1$
$k > 0$
$f''(x) = k(6x - 12)$
$f''(1) = k(6.1 - 12) = -6k < 0$ → $f(x)$ maksimum pada $x = 1$. $f''(3) = k(6.3 - 12) = 6k > 0$ → $f(x)$ minimum pada $x = 3$, berarti $a = 3$.
$\displaystyle \int_0^a f(x)dx = 27$
$\displaystyle \int_0^3 k(x^3 - 6x^2 + 9x)dx = 27$
$k\left(\dfrac14x^4 - 2x^3 + \dfrac92x^2\right)\Bigr|_0^3 = 27$
$k\left[\dfrac14.3^4 - 2.3^3 + \dfrac92.3^2 - \left(\dfrac14.0^4 - 2.0^3 + \dfrac92.0^2\right)\right] = 27$
$k\left[\dfrac{81}{4} - 54 + \dfrac{81}{2} - 0\right] = 27$
$\dfrac{27}{4}k = 27$
$k = 4$
jawab: D.
$f'(x) = 0$
$k(3x^2 - 12x + 9) = 0$
$3k(x^2 - 4x + 3) = 0$
$x^2 - 4x + 3 = 0$
$(x - 3)(x - 1) = 0$
$x = 3\ atau\ x = 1$
$k > 0$
$f''(x) = k(6x - 12)$
$f''(1) = k(6.1 - 12) = -6k < 0$ → $f(x)$ maksimum pada $x = 1$. $f''(3) = k(6.3 - 12) = 6k > 0$ → $f(x)$ minimum pada $x = 3$, berarti $a = 3$.
$\displaystyle \int_0^a f(x)dx = 27$
$\displaystyle \int_0^3 k(x^3 - 6x^2 + 9x)dx = 27$
$k\left(\dfrac14x^4 - 2x^3 + \dfrac92x^2\right)\Bigr|_0^3 = 27$
$k\left[\dfrac14.3^4 - 2.3^3 + \dfrac92.3^2 - \left(\dfrac14.0^4 - 2.0^3 + \dfrac92.0^2\right)\right] = 27$
$k\left[\dfrac{81}{4} - 54 + \dfrac{81}{2} - 0\right] = 27$
$\dfrac{27}{4}k = 27$
$k = 4$
jawab: D.
34. Soal Integral SNMPTN MtkIPA 2009
Jika pada $\displaystyle \int_{-1}^{2} x^2\sqrt{x + 1}\ dx$ disubstitusikan $u = x + 1$, maka menghasilkan . . . .
$A.\ \displaystyle \int_0^2 (u - 1)^2\sqrt{u}\ du$
$B.\ \displaystyle \int_0^1 (u - 1)^2\sqrt{u}\ du$
$C.\ \displaystyle \int_0^1 (x - 1)^2\sqrt{x}\ dx$
$D.\ \displaystyle \int_0^3 (u - 1)^2\sqrt{u}\ du$
$E.\ \displaystyle \int_0^3 (x - 1)^2\sqrt{x}\ dx$
$A.\ \displaystyle \int_0^2 (u - 1)^2\sqrt{u}\ du$
$B.\ \displaystyle \int_0^1 (u - 1)^2\sqrt{u}\ du$
$C.\ \displaystyle \int_0^1 (x - 1)^2\sqrt{x}\ dx$
$D.\ \displaystyle \int_0^3 (u - 1)^2\sqrt{u}\ du$
$E.\ \displaystyle \int_0^3 (x - 1)^2\sqrt{x}\ dx$
$u = x + 1 → x = u - 1$
$du = dx$
$x = -1 → u = 0;\ x = 2 → u = 3$
$\displaystyle \int_{-1}^{2} x^2\sqrt{x + 1}\ dx = \displaystyle \int_{0}^{3} (u - 1)^2\sqrt{u}\ du$
Misalkan:
$x = u → dx = du$
$u = 0 → x = 0; u = 3 → x = 3$
$\displaystyle \int_{0}^{3} (u - 1)^2\sqrt{u}\ du = \displaystyle \int_{0}^{3} (x - 1)^2\sqrt{x}\ dx$
jawab: E.
$du = dx$
$x = -1 → u = 0;\ x = 2 → u = 3$
$\displaystyle \int_{-1}^{2} x^2\sqrt{x + 1}\ dx = \displaystyle \int_{0}^{3} (u - 1)^2\sqrt{u}\ du$
Misalkan:
$x = u → dx = du$
$u = 0 → x = 0; u = 3 → x = 3$
$\displaystyle \int_{0}^{3} (u - 1)^2\sqrt{u}\ du = \displaystyle \int_{0}^{3} (x - 1)^2\sqrt{x}\ dx$
jawab: E.
35. Soal Integral SNMPTN MtkIPA 2009
Jika nilai $\displaystyle \int_1^2 f(x)dx = 6$, maka nilai $\displaystyle \int_0^1 xf(x^2 + 1)dx$ adalah . . . .
A. 1
B. 3
C. 4
D. 5
E. 6
A. 1
B. 3
C. 4
D. 5
E. 6
Misalkan:
$u = x^2 + 1$
$\dfrac{du}{dx} = 2x$
$\dfrac12du = xdx$
$x = 0 → u = 1;\ x = 1 → u = 2$
$\displaystyle \int_0^1 xf(x^2 + 1)dx = \displaystyle \int_1^2 f(u).\dfrac12du$
$= \displaystyle \dfrac12\int_1^2 f(u)du$
$= \displaystyle \dfrac12\int_1^2 f(x)dx$
$= \dfrac12.6$
$= 3$
jawab: B.
$u = x^2 + 1$
$\dfrac{du}{dx} = 2x$
$\dfrac12du = xdx$
$x = 0 → u = 1;\ x = 1 → u = 2$
$\displaystyle \int_0^1 xf(x^2 + 1)dx = \displaystyle \int_1^2 f(u).\dfrac12du$
$= \displaystyle \dfrac12\int_1^2 f(u)du$
$= \displaystyle \dfrac12\int_1^2 f(x)dx$
$= \dfrac12.6$
$= 3$
jawab: B.
36. Soal Integral SNMPTN MtkIPA
Jika $\displaystyle \int_1^2 \dfrac{1}{\sqrt{x} + 1}dx = a$, maka $\displaystyle \int_1^2 \dfrac{4\sqrt{x} + k}{\sqrt{x} + 1}dx = 4 - 3a$ untuk $k = \cdots$
$A.\ -3$
$B.\ -2$
$C.\ -1$
$D.\ 1$
$E.\ 2$
$A.\ -3$
$B.\ -2$
$C.\ -1$
$D.\ 1$
$E.\ 2$
$\displaystyle \int_1^2 \dfrac{4\sqrt{x} + k}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle \int_1^2 \dfrac{4\sqrt{x} + 4 + k - 4}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle \int_1^2 \dfrac{4\sqrt{x} + 4}{\sqrt{x} + 1}dx + \displaystyle \int_1^2 \dfrac{k - 4}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle 4\int_1^2 \dfrac{\sqrt{x} + 1}{\sqrt{x} + 1}dx + \displaystyle (k - 4)\int_1^2 \dfrac{1}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle 4\int_1^2 dx + (k - 4)a = 4 - 3a$
$4(x\bigr|_1^2) + (k - 4)a = 4 - 3a$
$4(2 - 1) + ak - 4a = 4 - 3a$
$4 + ak - 4a = 4 - 3a$
$ak = a$
$k = 1$
jawab: D.
$\displaystyle \int_1^2 \dfrac{4\sqrt{x} + 4 + k - 4}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle \int_1^2 \dfrac{4\sqrt{x} + 4}{\sqrt{x} + 1}dx + \displaystyle \int_1^2 \dfrac{k - 4}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle 4\int_1^2 \dfrac{\sqrt{x} + 1}{\sqrt{x} + 1}dx + \displaystyle (k - 4)\int_1^2 \dfrac{1}{\sqrt{x} + 1}dx = 4 - 3a$
$\displaystyle 4\int_1^2 dx + (k - 4)a = 4 - 3a$
$4(x\bigr|_1^2) + (k - 4)a = 4 - 3a$
$4(2 - 1) + ak - 4a = 4 - 3a$
$4 + ak - 4a = 4 - 3a$
$ak = a$
$k = 1$
jawab: D.
37. Soal Integral SIMAK UI MtkIPA 2014
Diberikan fungsi $f$ dan $g$ yang memenuhi sistem
$$\begin{cases}\displaystyle \int_0^1 f(x)dx + \left(\displaystyle \int_0^2 g(x)dx\right)^2 = 3 \\ f(x) = 3x^2 + 4x + \displaystyle \int_0^2 g(x)dx, \end{cases}$$
dengan $\displaystyle \int_0^2 g(x)dx \ne 0$. Nilai $f(1) = \cdots$
$A.\ -6$
$B.\ -3$
$C.\ 0$
$D.\ 3$
$E.\ 6$
$$\begin{cases}\displaystyle \int_0^1 f(x)dx + \left(\displaystyle \int_0^2 g(x)dx\right)^2 = 3 \\ f(x) = 3x^2 + 4x + \displaystyle \int_0^2 g(x)dx, \end{cases}$$
dengan $\displaystyle \int_0^2 g(x)dx \ne 0$. Nilai $f(1) = \cdots$
$A.\ -6$
$B.\ -3$
$C.\ 0$
$D.\ 3$
$E.\ 6$
Misalkan:
$\displaystyle \int_0^2 g(x)dx = p$
$f(x) = 3x^2 + 4x + p$
$\displaystyle \int_0^1 f(x)dx + \left(\displaystyle \int_0^2 g(x)dx\right)^2 = 3$
$\displaystyle \int_0^1 (3x^2 + 4x + p)dx + p^2 = 3$
$(x^3 + 2x^2 + px)\Bigr|_0^1 + p^2 = 3$
$1^3 + 2.1^2 + p.1 - (0^3 + 2.0^2 + p.0) + p^2 = 3$
$1 + 2 + p - 0 + p^2 = 3$
$p^2 + p = 0$
$p(p + 1) = 0$
$p = 0\ (tms)\ atau\ p = -1$
$\displaystyle \int_0^2 g(x)dx = p$
$\displaystyle \int_0^2 g(x)dx = -1$
$f(x) = 3x^2 + 4x + \displaystyle \int_0^2 g(x)dx$
$f(x) = 3x^2 + 4x - 1$
$f(1) = 3.1^2 + 4.1 - 1 = 6$
jawab: E.
$\displaystyle \int_0^2 g(x)dx = p$
$f(x) = 3x^2 + 4x + p$
$\displaystyle \int_0^1 f(x)dx + \left(\displaystyle \int_0^2 g(x)dx\right)^2 = 3$
$\displaystyle \int_0^1 (3x^2 + 4x + p)dx + p^2 = 3$
$(x^3 + 2x^2 + px)\Bigr|_0^1 + p^2 = 3$
$1^3 + 2.1^2 + p.1 - (0^3 + 2.0^2 + p.0) + p^2 = 3$
$1 + 2 + p - 0 + p^2 = 3$
$p^2 + p = 0$
$p(p + 1) = 0$
$p = 0\ (tms)\ atau\ p = -1$
$\displaystyle \int_0^2 g(x)dx = p$
$\displaystyle \int_0^2 g(x)dx = -1$
$f(x) = 3x^2 + 4x + \displaystyle \int_0^2 g(x)dx$
$f(x) = 3x^2 + 4x - 1$
$f(1) = 3.1^2 + 4.1 - 1 = 6$
jawab: E.
38. Soal Integral SIMAK UI MtkIPA 2014
Jika $\displaystyle \int_{-1}^{a} \dfrac{x + 1}{(x + 2)^4}dx = \dfrac{10}{81}$ dan $a > -2$, maka $a = \cdots$
$A.\ -1\dfrac12$
$B.\ -1$
$C.\ 0$
$D.\ 1$
$E.\ 1\dfrac12$
$A.\ -1\dfrac12$
$B.\ -1$
$C.\ 0$
$D.\ 1$
$E.\ 1\dfrac12$
$\displaystyle \int_{-1}^{a} \dfrac{x + 1}{(x + 2)^4}dx = \dfrac{10}{81}$
$\displaystyle \int_{-1}^{a} \dfrac{x + 2 - 1}{(x + 2)^4}dx = \dfrac{10}{81}$
$\displaystyle \int_{-1}^{a} \dfrac{x + 2}{(x + 2)^4}dx - \displaystyle \int_{-1}^{a} \dfrac{1}{(x + 2)^4}dx = \dfrac{10}{81}$
$\displaystyle \int_{-1}^{a} (x + 2)^{-3}dx - \displaystyle \int_{-1}^{a} (x + 2)^{-4}dx = \dfrac{10}{81}$
$\dfrac{1}{-3 + 1}(x + 2)^{-3 + 1}\Bigr|_{-1}^{a}$ $- \dfrac{1}{-4 + 1}(x + 2)^{-4 + 1}\Bigr|_{-1}^{a} = \dfrac{10}{81}$
$-\dfrac{1}{2}(x + 2)^{-2}\Bigr|_{-1}^{a} + \dfrac{1}{3}(x + 2)^{-3}\Bigr|_{-1}^{a} = \dfrac{10}{81}$
$-\dfrac12[(a + 2)^{-2} - (-1 + 2)^{-2}] +$ $\dfrac13[(a + 2)^{-3} - (-1 + 2)^{-3}] = \dfrac{10}{81}$
$-\dfrac12[(a + 2)^{-2} - 1] +$ $\dfrac13[(a + 2)^{-3} - 1] = \dfrac{10}{81}$
Misalkan:
$p = a + 2$
$-\dfrac12[p^{-2} - 1] + \dfrac13[p^{-3} - 1] = \dfrac{10}{81}$
$-\dfrac{1}{2p^2} + \dfrac12 + \dfrac{1}{3p^3} - \dfrac13 = \dfrac{10}{81}$ → semua dikali 6 !
$-\dfrac{3p}{p^3} + 3 + \dfrac{2}{p^3} - 2 = \dfrac{60}{81}$
$-\dfrac{3p}{p^3} + \dfrac{2}{p^3} = -\dfrac{21}{81}$
$\dfrac{3p - 2}{p^3} = \dfrac{7}{27}$
$3p - 2 = 7$
$3p = 9$
$p = 3$
$p = a + 2$
$3 = a + 2$
$a = 1$
jawab: D.
$\displaystyle \int_{-1}^{a} \dfrac{x + 2 - 1}{(x + 2)^4}dx = \dfrac{10}{81}$
$\displaystyle \int_{-1}^{a} \dfrac{x + 2}{(x + 2)^4}dx - \displaystyle \int_{-1}^{a} \dfrac{1}{(x + 2)^4}dx = \dfrac{10}{81}$
$\displaystyle \int_{-1}^{a} (x + 2)^{-3}dx - \displaystyle \int_{-1}^{a} (x + 2)^{-4}dx = \dfrac{10}{81}$
$\dfrac{1}{-3 + 1}(x + 2)^{-3 + 1}\Bigr|_{-1}^{a}$ $- \dfrac{1}{-4 + 1}(x + 2)^{-4 + 1}\Bigr|_{-1}^{a} = \dfrac{10}{81}$
$-\dfrac{1}{2}(x + 2)^{-2}\Bigr|_{-1}^{a} + \dfrac{1}{3}(x + 2)^{-3}\Bigr|_{-1}^{a} = \dfrac{10}{81}$
$-\dfrac12[(a + 2)^{-2} - (-1 + 2)^{-2}] +$ $\dfrac13[(a + 2)^{-3} - (-1 + 2)^{-3}] = \dfrac{10}{81}$
$-\dfrac12[(a + 2)^{-2} - 1] +$ $\dfrac13[(a + 2)^{-3} - 1] = \dfrac{10}{81}$
Misalkan:
$p = a + 2$
$-\dfrac12[p^{-2} - 1] + \dfrac13[p^{-3} - 1] = \dfrac{10}{81}$
$-\dfrac{1}{2p^2} + \dfrac12 + \dfrac{1}{3p^3} - \dfrac13 = \dfrac{10}{81}$ → semua dikali 6 !
$-\dfrac{3p}{p^3} + 3 + \dfrac{2}{p^3} - 2 = \dfrac{60}{81}$
$-\dfrac{3p}{p^3} + \dfrac{2}{p^3} = -\dfrac{21}{81}$
$\dfrac{3p - 2}{p^3} = \dfrac{7}{27}$
$3p - 2 = 7$
$3p = 9$
$p = 3$
$p = a + 2$
$3 = a + 2$
$a = 1$
jawab: D.
39. Soal Integral SIMAK UI 2013 MtkIPA
$\displaystyle \int_0^2 \dfrac{x^2 + 3x}{\sqrt{x + 2}}dx = \dots$
$A.\ \dfrac{4}{15}(7 - \sqrt{2})$
$B.\ \dfrac{4}{15}(7\sqrt{2} - 1)$
$C.\ \dfrac{4}{15}(7\sqrt{2} + 1)$
$D.\ \dfrac{8}{15}(7\sqrt{2} - 1)$
$E.\ \dfrac{8}{15}(7\sqrt{2} + 1)$
$A.\ \dfrac{4}{15}(7 - \sqrt{2})$
$B.\ \dfrac{4}{15}(7\sqrt{2} - 1)$
$C.\ \dfrac{4}{15}(7\sqrt{2} + 1)$
$D.\ \dfrac{8}{15}(7\sqrt{2} - 1)$
$E.\ \dfrac{8}{15}(7\sqrt{2} + 1)$
Misalkan:
$u = x + 2 → x = u - 2$
$du = dx$
$x = 0 → u = 2;\ x = 2 → u = 4$
$\displaystyle \int_0^2 \dfrac{x^2 + 3x}{\sqrt{x + 2}}dx = \displaystyle \int_2^4 \dfrac{(u - 2)^2 + 3(u - 2)}{\sqrt{u}}du$
$= \displaystyle \int_2^4 \dfrac{u^2 - 4u + 4 + 3u - 6}{\sqrt{u}}du$
$= \displaystyle \int_2^4 \dfrac{u^2 - u - 2}{\sqrt{u}}du$
$= \displaystyle \int_2^4 (u^{3/2} - u^{1/2} - 2u^{-1/2})du$
$= (\dfrac25u^{5/2} - \dfrac23u^{3/2} - 4u^{1/2})\Bigr|_2^4$
$= \dfrac25.4^{5/2} - \dfrac23.4^{3/2} - 4.4^{1/2} - \left(\dfrac25.2^{5/2} - \dfrac23.2^{3/2} - 4.2^{1/2}\right)$
$= \dfrac25.32 - \dfrac23.8 - 8 - \left(\dfrac85\sqrt{2} - \dfrac43\sqrt{2} - 4\sqrt{2}\right)$
$= \dfrac{8}{15} - \left(-\dfrac{56}{15}\sqrt{2}\right)$
$= \dfrac{8}{15} + \dfrac{56}{15}\sqrt{2}$
$= \dfrac{8}{15}\left(7\sqrt{2} + 1\right)$
jawab: E.
$u = x + 2 → x = u - 2$
$du = dx$
$x = 0 → u = 2;\ x = 2 → u = 4$
$\displaystyle \int_0^2 \dfrac{x^2 + 3x}{\sqrt{x + 2}}dx = \displaystyle \int_2^4 \dfrac{(u - 2)^2 + 3(u - 2)}{\sqrt{u}}du$
$= \displaystyle \int_2^4 \dfrac{u^2 - 4u + 4 + 3u - 6}{\sqrt{u}}du$
$= \displaystyle \int_2^4 \dfrac{u^2 - u - 2}{\sqrt{u}}du$
$= \displaystyle \int_2^4 (u^{3/2} - u^{1/2} - 2u^{-1/2})du$
$= (\dfrac25u^{5/2} - \dfrac23u^{3/2} - 4u^{1/2})\Bigr|_2^4$
$= \dfrac25.4^{5/2} - \dfrac23.4^{3/2} - 4.4^{1/2} - \left(\dfrac25.2^{5/2} - \dfrac23.2^{3/2} - 4.2^{1/2}\right)$
$= \dfrac25.32 - \dfrac23.8 - 8 - \left(\dfrac85\sqrt{2} - \dfrac43\sqrt{2} - 4\sqrt{2}\right)$
$= \dfrac{8}{15} - \left(-\dfrac{56}{15}\sqrt{2}\right)$
$= \dfrac{8}{15} + \dfrac{56}{15}\sqrt{2}$
$= \dfrac{8}{15}\left(7\sqrt{2} + 1\right)$
jawab: E.
40. Soal Integral SIMAK UI Mtk IPA 2013
Jika dari persamaan
$\displaystyle \int_{-1}^{x}(12t^2 + 40t + 29)dt = -4(^{x + 2}log\sqrt{x + 2}) - 2x$ diperoleh bentuk $F(x) = 0$, maka sisa pembagian $F(x)$ oleh $x^2 + 2x + 2$ adalah . . . .
$A.\ x + 9$
$B.\ 4x + 12$
$C.\ -x - 9$
$D.\ x + 3$
$E.\ -x - 3$
$\displaystyle \int_{-1}^{x}(12t^2 + 40t + 29)dt = -4(^{x + 2}log\sqrt{x + 2}) - 2x$ diperoleh bentuk $F(x) = 0$, maka sisa pembagian $F(x)$ oleh $x^2 + 2x + 2$ adalah . . . .
$A.\ x + 9$
$B.\ 4x + 12$
$C.\ -x - 9$
$D.\ x + 3$
$E.\ -x - 3$
$\displaystyle \int_{-1}^{x}(12t^2 + 40t + 29)dt = -4(^{x + 2}log\sqrt{x + 2}) - 2x$
$(4t^3 + 20t^2 + 29t)\Bigr|_{-1}^{x} = -4.\dfrac12.(^{x + 2}log\ (x + 2)) - 2x$
$4x^3 + 20x^2 + 29x - (4.(-1)^3 + 20.(-1)^2 + 29.(-1))$ $= -2 - 2x$
$4x^3 + 20x^2 + 29x - (-13) = -2 - 2x$
$4x^3 + 20x^2 + 31x + 15 = 0$
Lakukan pembagian langsung !
jawab: C.
$(4t^3 + 20t^2 + 29t)\Bigr|_{-1}^{x} = -4.\dfrac12.(^{x + 2}log\ (x + 2)) - 2x$
$4x^3 + 20x^2 + 29x - (4.(-1)^3 + 20.(-1)^2 + 29.(-1))$ $= -2 - 2x$
$4x^3 + 20x^2 + 29x - (-13) = -2 - 2x$
$4x^3 + 20x^2 + 31x + 15 = 0$
Lakukan pembagian langsung !
Demikianlah soal dan pembahasan integral fungsi aljabar, semoga bermanfaat. Selamat belajar !
Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB
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