Integral Tak Tentu Bersama Sifat-Sifatnya Beserta Contoh Soalnya

Contoh Soal Pilihan Ganda dan Pembahasannya yang Berkaitan dengan Integral Tak Tentu

DEFINISI INTEGRAL FUNGSI TAK TENTU

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Ketika akan menyelesaikan persamaan diferensial dari bentuk dydx=f(x) dapat kita tulis dalam bentuk dy=f(x)dx. Secara umum, jika F(x) menyatakan fungsi dalam variabel x, dengan f(x) turunan dari F(x) dan c konstanta bilangan real maka integral tak tentu dari f(x) dapat dituliskan dalam bentuk: ∫f(x)dx=F(x)+c
dibaca:"integral fungsi f(x) ke x sama dengan F(x)+c"

Keterangan Tambahan:

∫f(x)=notasi integral tak tentu
F(x)+c=fungsi antiturunan
f(x)=fungsi yang diintegralkan (integran)
c=konstanta
d(x)=diferensial (turunan) dari x

ATURAN DASAR INTEGRAL TAK TENTU

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• ∫dx=x+c
• ∫k dx=kx+c
• ∫xn dx=1n+1xn+c, n≠−1
• ∫kf(x) dx=k∫f(x)dx
• ∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx
• ∫[f(x)−g(x)]dx=∫f(x)dx−∫g(x)dx
• ∫axdx=(1ln a)ax+c
• ∫au(x)dx=(1u′(x) ln a)au(x)+c
• ∫1xdx=ln |x|+c
• ∫1u(x)dx=1u′(x)ln |u(x)|+c
• ∫exdx=ex+c
• ∫eu(x)dx=1u′(x)eu(x)+c

1. Soal UNBK Matematika IPA 2019 (*Soal Lengkap)

$\int \left(12x^2-4x+1\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& 6x^3-4x^2+x+C\\ \left(B\right)& 6x^3-4x^2+C\\ \left(C\right)& 4x^3+2x^2+x+C\\ \left(D\right)& 4x^3-2x^2+x+C\\ \left(E\right)& 4x^3+2x^2+x+C\end{array}$

Alternatif Pembahasan:

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Dengan menerapkan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \left(12x^2-4x+1\right)dx\\ & =\frac{12}{2+1}{x}^{2+1}-\frac{4}{1+1}{x}^{1+1}+1x+C\\ & =4x^3-2x^2+x+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)4x^3-2x^2+x+C$

2. Soal UNBK Matematika IPA 2019 (*Soal Lengkap)

Hasil dari $\int \left(3x^2-5x+4\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& x^3-\frac{5}{2}{x}^2+4x+C\\ \left(B\right)& x^3-5x^2+4x+C\\ \left(C\right)& 3x^3-5x^2+4x+C\\ \left(D\right)& 6x^3-5x^2+4x+C\\ \left(E\right)& 6x^3-\frac{5}{2}{x}^2+4x+C\end{array}$

Alternatif Pembahasan:

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Dengan menerapkan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \left(3x^2-5x+4\right)dx\\ & =\frac{3}{2+1}{x}^{2+1}-\frac{5}{1+1}{x}^{1+1}+4x+C\\ & =x^3-\frac{5}{2}{x}^2+4x+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(A\right)x^3-\frac{5}{2}{x}^2+4x+C$

3. Soal UNBK Matematika SMA IPS 2019 (*Soal Lengkap)

Hasil dari $\int \left(2x^3-9x^2+4x-5\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& \frac{1}{2}{x}^4-6x^3+2x^2-5x+C\\ \left(B\right)& \frac{1}{2}{x}^4-6x^3+x^2-5x+C\\ \left(C\right)& \frac{1}{2}{x}^4-3x^3+x^2-5x+C\\ \left(D\right)& \frac{1}{2}{x}^4-3x^3+2x^2-5x+C\\ \left(E\right)& \frac{1}{2}{x}^4-6x^3-2x^2-5x+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \left(2x^3-9x^2+4x-5\right)\\ & =\frac{2}{3+1}{x}^{3+1}-\frac{9}{2+1}{x}^{2+1}+\frac{4}{1+1}{x}^{1+1}-5x+C\\ & =\frac{2}{4}{x}^4-\frac{9}{3}{x}^3+\frac{4}{2}{x}^2-5x+C\\ & =\frac{1}{2}{x}^4-3x^3+2x^2-5x+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)\frac{1}{2}{x}^4-3x^3+2x^2-5x+C$

4. Soal UNBK Matematika IPA 2019 (*Soal Lengkap)

Hasil dari $\int \left(x-2\right){\left(x^2-4x+3\right)}^{5}dx$ adalah...
$\begin{array}{cc}\left(A\right)& \frac{1}{3}{\left(x^2-4x+3\right)}^6+C\\ \left(B\right)& \frac{1}{6}{\left(x^2-4x+3\right)}^6+C\\ \left(C\right)& \frac{1}{12}{\left(x^2-4x+3\right)}^6+C\\ \left(D\right)& \frac{1}{6}{\left(x-2\right)}^2{\left(x^2-4x+3\right)}^6+C\\ \left(E\right)& \frac{1}{6}{\left(x-2\right)}^2{\left(x^2-4x+3\right)}^6+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
misal:
$\begin{array}{cc}u& =x^2-4x+3\\ \frac{du}{dx}& =2x-4\\ \frac{du}{dx}& =2(x-2)\\ \frac{1}{2}du& =(x-2)dx\end{array}$

Soal di atas, kini bisa kita tulis menjadi;
$\begin{array}{cc}& \int \left(x-2\right){\left(x^2-4x+3\right)}^{5}dx\\ & =\int u^5\left(x-2\right)dx\\ & =\int u^5\cdot \frac{1}{2}du\\ & =\frac{1}{5+1}{u}^{5+1}\cdot \frac{1}{2}+C\\ & =\frac{1}{12}{u}^6+C\\ & =\frac{1}{12}{\left(x^2-4x+3\right)}^6+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(C\right)\frac{1}{12}{\left(x^2-4x+3\right)}^6+C$

5. Soal UNBK Matematika IPA 2019 (*Soal Lengkap)

Hasil dari $\int \left(2x-1\right){\left(x^2-x+3\right)}^{3}dx$ adalah...
$\begin{array}{cc}\left(A\right)& \frac{1}{3}{\left(x^2-x+3\right)}^3+C\\ \left(B\right)& \frac{1}{4}{\left(x^2-x+3\right)}^3+C\\ \left(C\right)& \frac{1}{4}{\left(x^2-x+3\right)}^4+C\\ \left(D\right)& \frac{1}{2}{\left(x^2-x+3\right)}^4+C\\ \left(E\right)& {\left(x^2-x+3\right)}^4+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
misal:
$\begin{array}{cc}u& =x^2-x+3\\ \frac{du}{dx}& =2x-1\\ du& =(2x-1)dx\end{array}$

Soal di atas, kini bisa kita tulis menjadi;
$\begin{array}{cc}& \int \left(2x-1\right){\left(x^2-x+3\right)}^{3}dx\\ & =\int {\left(x^2-x+3\right)}^3\left(2x-1\right)dx\\ & =\int {\left(u\right)}^{3}du\\ & =\frac{1}{3+1}{u}^{3+1}+C\\ & =\frac{1}{4}{\left(x^2-x+3\right)}^4+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(C\right)\frac{1}{4}{\left(x^2-x+3\right)}^4+C$

6. Soal UM STIS 2013 (*Soal Lengkap)

Hasil dari $\int x\sqrt{4x+1}dx$ adalah...
$\begin{array}{cc}\left(A\right)& -\frac{1}{60}\left(6x-1\right){\left(4x+1\right)}^{\frac{3}{2}}+C\\ \left(B\right)& \frac{1}{60}\left(6x-1\right){\left(4x+1\right)}^{\frac{3}{2}}+C\\ \left(C\right)& -\frac{4}{60}\left(3x+2\right){\left(4x+1\right)}^{\frac{3}{2}}+C\\ \left(D\right)& \frac{4}{60}\left(3x+2\right){\left(4x+1\right)}^{\frac{3}{2}}+C\\ \left(E\right)& \frac{1}{60}\left(3x+2\right){\left(4x+1\right)}^{\frac{3}{2}}+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}u& =4x+1\to x=\frac{1}{4}\left(u-1\right)\\ du& =4dx\\ \frac{1}{4}du& =dx\end{array}$
Apa yang kita peroleh di atas, kita coba substituskan ke soal dan kita lakukan manipulasi aljabar;

$\begin{array}{cc}\int x\sqrt{4x+1}dx& =\int \frac{1}{4}\left(u-1\right)\sqrt{u}dx\\ & =\int \frac{1}{4}\left(u-1\right)\cdot u^{\frac{1}{2}}\frac{1}{4}du\\ & =\frac{1}{4}\cdot \frac{1}{4}\int \left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right)du\\ & =\frac{1}{4}\cdot \frac{1}{4}\left[\frac{2}{5}{u}^{\frac{5}{2}}-\frac{2}{3}{u}^{\frac{3}{2}}\right]+C\\ & =\frac{1}{4}\cdot \left(\frac{1}{10}{u}^{\frac{5}{2}}-\frac{1}{6}{u}^{\frac{3}{2}}\right)+C\\ & =\frac{1}{4}\cdot \frac{1}{60}{u}^{\frac{3}{2}}\left(6u^1-10\right)+C\\ & =\frac{1}{4}\cdot \frac{1}{60}{\left(4x+1\right)}^{\frac{3}{2}}\left(6\left(4x+1\right)-10\right)+C\\ & =\frac{1}{4}\cdot \frac{1}{60}{\left(4x+1\right)}^{\frac{3}{2}}\left(24x+6-10\right)+C\\ & =\frac{1}{4}\cdot \frac{1}{60}{\left(4x+1\right)}^{\frac{3}{2}}\left(24x-4\right)+C\\ & =\frac{1}{60}{\left(4x+1\right)}^{\frac{3}{2}}\left(6x-1\right)+C\end{array}$

$\therefore$ Pilihan yang sesuai $\left(B\right)\frac{1}{60}\left(6x-1\right){\left(4x+1\right)}^{\frac{3}{2}}+C$

7. Soal SBMPTN 2018 Kode 526 (*Soal Lengkap)

Hasil dari $\int {\left(2x-\frac{1}{2x}\right)}^{2}dx$ adalah...
$\begin{array}{cc}\left(A\right)& \frac{2}{3}{x}^3-\frac{1}{2x}-2x+C\\ \left(B\right)& \frac{2}{3}{x}^3+\frac{1}{2x}-2x+C\\ \left(C\right)& \frac{4}{3}{x}^3-\frac{1}{2x}+2x+C\\ \left(D\right)& \frac{4}{3}{x}^3-\frac{1}{4x}-2x+C\\ \left(E\right)& \frac{4}{3}{x}^3+\frac{1}{4x}-2x+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int {\left(2x-\frac{1}{2x}\right)}^{2}dx\\ & =\int \left(4x^2-2+\frac{1}{4x^2}\right)dx\\ & =\int \left(4x^2-2+\frac{1}{4}{x}^{-2}\right)dx\\ & =\frac{4}{2+1}{x}^{2+1}-2x+\frac{\frac{1}{4}}{-2+1}{x}^{-2+1}+C\\ & =\frac{4}{3}{x}^3-2x-\frac{1}{4x}+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)\frac{4}{3}{x}^3-\frac{1}{4x}-2x+C$

8. Soal UMPTN 1995 Rayon C (*Soal Lengkap)

Diketahui $f\left(x\right)=\int x^{2}dx$. Jika $f\left(2\right)=-\frac{19}{3}$, maka kurva itu memotong sumbu $x$ pada...

$\begin{array}{cc}\left(A\right)& \left(0,0\right)\\ \left(B\right)& \left(1,0\right)\\ \left(C\right)& \left(2,0\right)\\ \left(D\right)& \left(3,0\right)\\ \left(E\right)& \left(4,0\right)\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}f\left(x\right)& =\int x^{2}dx\\ & =\frac{1}{2+1}{x}^{2+1}+c\\ & =\frac{1}{3}{x}^3+c\\ f\left(2\right)& =\frac{1}{3}(2{)}^3+c\\ -\frac{19}{3}& =\frac{8}{3}+c\\ -\frac{19}{3}-\frac{8}{3}& =c\\ -\frac{27}{3}& =c\\ -9& =c\\ f\left(x\right)& =\frac{1}{3}{x}^3+c\\ & =\frac{1}{3}{x}^3-9\end{array}$

$f\left(x\right)=\frac{1}{3}{x}^3-9$ memotong sumbu $x$ saat $\frac{1}{3}{x}^3-9=0$ sehingga $x^3-27=0$ atau $x=3$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)\left(3,0\right)$

9. Soal SNMPTN 2011 Kode 678 (*Soal Lengkap)

Diketahui $\int f\left(x\right)dx=\frac{1}{4}a{x}^2+bx+c$ dan $a\ne 0$. Jika $f\left(a\right)=\frac{a+2b}{2}$ dan $f\left(b\right)=6$, maka fungsi $f\left(x\right)=\cdots$

$\begin{array}{cc}\left(A\right)& \frac{1}{2}x+4\\ \left(B\right)& 2x+4\\ \left(C\right)& \frac{1}{2}x-4\\ \left(D\right)& x+4\\ \left(E\right)& -\frac{1}{2}x+4\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:

Untuk $\int f\left(x\right)dx=\frac{1}{4}a{x}^2+bx+c$ dapat kita tentukan $f\left(x\right)=\frac{1}{2}ax+b$

Untuk $f\left(a\right)=\frac{a+2b}{2}$ maka berlaku:
$\begin{array}{cc}f\left(x\right)& =\frac{1}{2}ax+b\\ f\left(a\right)& =\frac{1}{2}a\left(a\right)+b\\ \frac{a+2b}{2}& =\frac{1}{2}{a}^2+b\\ a+2b& =a^2+2b\\ 0& =a^2-a\\ 0& =a\left(a-1\right)\\ & a=0\text{atau}a=1\end{array}$

Nilai $a$ yang memenuhi adalah $a=1$, sehingga $f\left(x\right)=\frac{1}{2}x+b$. Untuk $f\left(b\right)=6$, kita peroleh:
$\begin{array}{cc}f\left(x\right)& =\frac{1}{2}x+b\\ f\left(b\right)& =\frac{1}{2}b+b\\ 6& =\frac{3}{2}b\\ b& =4\\ f\left(x\right)& =\frac{1}{2}x+4\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(A\right)\frac{1}{2}x+4$

10. Soal SNMPTN 2011 Kode 591 (*Soal Lengkap)

Diberikan $f\left(x\right)=ax+b$ dan $F\left(x\right)$ adalah anti turunan $f\left(x\right)$. Jika $F\left(1\right)-F\left(0\right)=3$ maka $2a+b$ adalah...

$\begin{array}{cc}\left(A\right)& 10\\ \left(B\right)& 6\\ \left(C\right)& 5\\ \left(D\right)& 4\\ \left(E\right)& 3\end{array}$

Alternatif Pembahasan:

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Untuk $f\left(x\right)=a+bx$ dan $F\left(x\right)$ adalah anti turunan $f\left(x\right)$ maka berlaku:

$\begin{array}{cc}F\left(x\right)& =\int f\left(x\right)dx\\ F\left(x\right)& =\int \left(a+bx\right)dx\\ F\left(x\right)& =ax+\frac{1}{2}b{x}^2+c\\ F\left(1\right)& =a\left(1\right)+\frac{1}{2}b(1{)}^2+c\\ F\left(1\right)& =a+\frac{1}{2}b+c\\ F\left(0\right)& =a\left(0\right)+\frac{1}{2}b(0{)}^2+c\\ F\left(0\right)& =c\\ F\left(1\right)-F\left(0\right)& =a+\frac{1}{2}b+c-c\\ 3& =a+\frac{1}{2}b\\ 6& =2a+b\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)6$

11. Soal SBMPTN 2018 Kode 527 (*Soal Lengkap)

Hasil dari $\int \left(\frac{-16-6x^4}{x^2}\right)dx$ adalah...
$\begin{array}{cc}\left(A\right)& \frac{16}{x}+2x^3+C\\ \left(B\right)& \frac{16}{x}-2x^3+C\\ \left(C\right)& -\frac{16}{x}-x^3+C\\ \left(D\right)& -\frac{8}{x}+2x^3+C\\ \left(E\right)& \frac{8}{x}-2x^3+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \left(\frac{-16-6x^4}{x^2}\right)dx\\ & =\int \left(\frac{-16}{x^2}-\frac{6x^4}{x^2}\right)dx\\ & =\int \left(-16x^{-2}-6x^{4-2}\right)dx\\ & =\int \left(-16x^{-2}-6x^2\right)dx\\ & =\frac{-16}{-2+1}{x}^{-2+1}-\frac{6}{2+1}{x}^{2+1}+C\\ & =16x^{-1}-2x^3+C\\ & =\frac{16}{x}-2x^3+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)\frac{16}{x}-2x^3+C$

12. Soal SBMPTN 2017 SAINTEK Kode 226 (*Soal Lengkap)

$\int \frac{3\left(1-x\right)}{1+\sqrt{x}}dx=\cdots$

$\begin{array}{cc}\left(A\right)& 3x-2x\sqrt{x}+C\\ \left(B\right)& 2x-3x\sqrt{x}+C\\ \left(C\right)& 3x\sqrt{x}-2x+C\\ \left(D\right)& 2x\sqrt{x}-3x+C\\ \left(E\right)& 3x+2x\sqrt{x}+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \frac{3\left(1-x\right)}{1+\sqrt{x}}dx\\ & =\int \frac{3\left(1-x\right)}{1+\sqrt{x}}\times \frac{1-\sqrt{x}}{1-\sqrt{x}}dx\\ & =\int \frac{3\left(1-x\right)\left(1-\sqrt{x}\right)}{1-x}dx\\ & =3\int \left(1-\sqrt{x}\right)dx\\ & =3\left(x-\frac{2}{3}x\sqrt{x}\right)+C\\ & =3x-2x\sqrt{x}+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(A\right)3x-2x\sqrt{x}+C$

13. Soal SBMPTN 2017 SAINTEK Kode 241 (*Soal Lengkap)

$\int \frac{x^2-\sqrt{x}}{x}dx=\cdots$

$\begin{array}{cc}\left(A\right)& x^2+2\sqrt{x}+C\\ \left(B\right)& \frac{1}{2}{x}^2+2\sqrt{x}+C\\ \left(C\right)& \frac{1}{2}{x}^2-2\sqrt{x}+C\\ \left(D\right)& \frac{1}{2}{x}^2+x\sqrt{x}+C\\ \left(E\right)& \frac{1}{2}{x}^2-x\sqrt{x}+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \frac{x^2-\sqrt{x}}{x}dx\\ & =\int \left(\frac{x^2}{x}-\frac{\sqrt{x}}{x}\right)dx\\ & =\int \left(x-x^{-\frac{1}{2}}\right)dx\\ & =\frac{1}{2}{x}^2-2x^{\frac{1}{2}}+C\\ & =\frac{1}{2}{x}^2-2\sqrt{x}+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(C\right)\frac{1}{2}{x}^2-2\sqrt{x}+C$

14. Soal SBMPTN 2017 SAINTEK Kode 229 (*Soal Lengkap)

$\int 9x^2\sqrt{x^3-1}dx=\cdots$

$\begin{array}{cc}\left(A\right)& \frac{1}{3}\left(x^3-1\right)\sqrt{x^3-1}+C\\ \left(B\right)& \left(x^3-1\right)\sqrt{x^3-1}+C\\ \left(C\right)& 2\left(x^3-1\right)\sqrt{x^3-1}+C\\ \left(D\right)& 3\left(x^3-1\right)\sqrt{x^3-1}+C\\ \left(E\right)& 9\left(x^3-1\right)\sqrt{x^3-1}+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:

misal:
$\begin{array}{cc}u& =x^3-1\\ \frac{du}{dx}& =3x^2\\ du& =3x^{2}dx\end{array}$
Soal di atas, kini dapat kita tuliskan menjadi;
$\begin{array}{cc}& \int 9x^2\sqrt{x^3-1}dx\\ & =\int 3\cdot 3x^2\sqrt{x^3-1}dx\\ & =\int 3\cdot \sqrt{x^3-1}3x^{2}dx\\ & =\int 3\cdot \sqrt{u}du\\ & =3\cdot \frac{2}{3}\cdot \left(u\right)\sqrt{u}+C\\ & =3\cdot \frac{2}{3}\cdot \left(x^3-1\right)\sqrt{x^3-1}+C\\ & =2\cdot \left(x^3-1\right)\sqrt{x^3-1}+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(C\right)2\left(x^3-1\right)\sqrt{x^3-1}+C$

15. Soal SBMPTN 2018 Kode 527 (*Soal Lengkap)

$\int \left(\frac{-16-6x^4}{x^2}\right)dx=\cdots$

$\begin{array}{cc}\left(A\right)& \frac{16}{x}+2x^3+C\\ \left(B\right)& \frac{16}{x}-2x^3+C\\ \left(C\right)& -\frac{16}{x}-2x^3+C\\ \left(D\right)& -\frac{8}{x}+2x^3+C\\ \left(E\right)& \frac{8}{x}-2x^3+C\end{array}$

Alternatif Pembahasan:

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Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int \left(\frac{-16-6x^4}{x^2}\right)dx\\ & =\int \left(\frac{-16}{x^2}-\frac{6x^4}{x^2}\right)dx\\ & =\int \left(-16x^{-2}-6x^2\right)dx\\ & =\frac{-16}{-2+1}{x}^{-2+1}-\frac{6}{2+1}{x}^{2+1}+C\\ & =16x^{-1}-2x^3+C\\ & =\frac{16}{x}-2x^3+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)\frac{16}{x}-2x^3+C$

16. Soal SBMPTN 2018 Kode 527 (*Soal Lengkap)

$\int {\left(\frac{x^4-1}{x^3+x}\right)}^{2}dx=\cdots$

$\begin{array}{cc}\left(A\right)& \frac{1}{3}{x}^3+\frac{1}{x}-2x+C\\ \left(B\right)& \frac{1}{3}{x}^3-\frac{1}{x}-2x+C\\ \left(C\right)& \frac{1}{3}{x}^3+\frac{1}{x}+2x+C\\ \left(D\right)& \frac{1}{3}{x}^3-\frac{1}{x}+x+C\\ \left(E\right)& \frac{1}{3}{x}^3-\frac{1}{x}-x+C\end{array}$

Alternatif Pembahasan:

Show

Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}& \int {\left(\frac{x^4-1}{x^3+x}\right)}^{2}dx\\ & =\int {\left(\frac{\left(x^2-1\right)\left(x^2+1\right)}{x\left(x^2+1\right)}\right)}^{2}dx\\ & =\int {\left(\frac{\left(x^2-1\right)}{x}\right)}^{2}dx\\ & =\int {\left(\frac{x^2}{x}-\frac{1}{x}\right)}^{2}dx\\ & =\int {\left(x-x^{-1}\right)}^{2}dx\\ & =\int \left(x^2-2+x^{-2}\right)dx\\ & =\frac{1}{2+1}{x}^{2+1}-2x+\frac{1}{-2+1}{x}^{-2+1}+C\\ & =\frac{1}{3}{x}^3-2x-\frac{1}{x}+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)\frac{1}{3}{x}^3-\frac{1}{x}-2x+C$

17. Soal UM UNDIP 2016 Kode 501/515 (*Soal Lengkap)

$\int x^5{\left(2-x^3\right)}^{\frac{1}{2}}dx=\cdots$

$\begin{array}{cc}\left(A\right)& \frac{2}{45}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C\\ \left(B\right)& \frac{-2}{5}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C\\ \left(C\right)& \frac{2}{5}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C\\ \left(D\right)& \frac{-2}{25}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C\\ \left(E\right)& \frac{-2}{45}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C\end{array}$

Alternatif Pembahasan:

Show

Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
misal:
$\begin{array}{cc}u& =2-x^3\to 2-u=x^3\\ \frac{du}{dx}& =-3x^2\\ du& =-3x^{2}dx\to -\frac{1}{3}du=x^{2}dx\end{array}$

Soal di atas, kini bisa kita tulis menjadi;
$\begin{array}{cc}& \int x^5{\left(2-x^3\right)}^{\frac{1}{2}}dx\\ & =\int x^2\cdot x^3{\left(u\right)}^{\frac{1}{2}}dx\\ & =\int x^3\cdot u^{\frac{1}{2}}{x}^{2}dx\\ & =\int \left(2-u\right)u^{\frac{1}{2}}\left(-\frac{1}{3}du\right)\\ & =-\frac{1}{3}\int \left(2u^{\frac{1}{2}}-u^{\frac{3}{2}}\right)du\\ & =-\frac{1}{3}\cdot \left(\frac{4}{3}{u}^{\frac{3}{2}}-\frac{2}{5}{u}^{\frac{5}{2}}\right)+C\\ & =-\frac{1}{3}\cdot \left(\frac{4}{3}{u}^{\frac{3}{2}}-\frac{2}{5}{u}^{\frac{5}{2}}\right)+C\\ & =-\frac{1}{3}\cdot \frac{1}{15}{u}^{\frac{3}{2}}\left(20-6u\right)+C\\ & =-\frac{1}{45}{\left(2-x^3\right)}^{\frac{3}{2}}\left(20-6\left(2-x^3\right)\right)+C\\ & =-\frac{1}{45}{\left(2-x^3\right)}^{\frac{3}{2}}\left(20-12+6x^3\right)+C\\ & =-\frac{1}{45}{\left(2-x^3\right)}^{\frac{3}{2}}\left(8+6x^3\right)+C\\ & =-\frac{2}{45}{\left(2-x^3\right)}^{\frac{3}{2}}\left(4+3x^3\right)+C\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(E\right)\frac{-2}{45}\left(3x^3+4\right){\left(-x^3+2\right)}^{\frac{3}{2}}+C$

18. Soal UM UNDIP 2016 Kode 501/515 (*Soal Lengkap)

Jika $f^{\prime }\left(x\right)=9x^2-12x+2$ dan $f(-1)=0$, maka $f\left(0\right)=\cdots$

$\begin{array}{cc}\left(A\right)& -1\\ \left(B\right)& 0\\ \left(C\right)& 9\\ \left(D\right)& 11\\ \left(E\right)& 13\end{array}$

Alternatif Pembahasan:

Show

Untuk menyelesaikan soal Integral di atas kita coba dengan menggunakan aturan dasar integral $\int x^{n}dx=\frac{1}{n+1}{x}^n+c,n\ne -1$ dan manipulasi aljabar, maka kita akan peroleh:
$\begin{array}{cc}{f}^{\prime }\left(x\right)& =9x^2-12x+2\\ f\left(x\right)& =\int f^{\prime }\left(x\right)dx\\ & =\int 9x^2-12x+2dx\\ & =\frac{9}{2+1}{x}^{2+1}-\frac{12}{1+1}{x}^{1+1}+2x+C\\ & =3x^3-6x^2+2x+C\\ f(-1)& =3(-1{)}^3-6(-1{)}^2+2(-1)+C\\ 0& =-3-6-2+C\\ 0& =-11+C\\ C& =11\\ f\left(x\right)& =3x^3-6x^2+2x+11\\ f\left(0\right)& =3(0{)}^3-6(0{)}^2+2\left(0\right)+11\\ & =11\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)11$

19. Soal UTBK Tes Kompetensi Akademik SAINTEK 2019

Jika nilai $\underset{b}{\overset{a}{\int }}f\left(x\right)dx=5$ dan $\underset{c}{\overset{a}{\int }}f\left(x\right)dx=0$, maka $\underset{c}{\overset{b}{\int }}f\left(x\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& -5\\ \left(B\right)& -3\\ \left(C\right)& 0\\ \left(D\right)& 4\\ \left(E\right)& 6\end{array}$

Alternatif Pembahasan:

Show

Catatan calon guru yang mungkin bermanfaat tentang sifat integral tentu;

• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=-\underset{b}{\overset{a}{\int }}f\left(x\right)dx$
• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx+\underset{b}{\overset{c}{\int }}f\left(x\right)dx=\underset{a}{\overset{c}{\int }}f\left(x\right)dx$

$\begin{array}{cc}\underset{b}{\overset{a}{\int }}f\left(x\right)dx=5& \Rightarrow \underset{a}{\overset{b}{\int }}f\left(x\right)dx=-5\\ \underset{c}{\overset{a}{\int }}f\left(x\right)dx=0& \Rightarrow \underset{a}{\overset{c}{\int }}f\left(x\right)dx=0\\ \underset{c}{\overset{b}{\int }}f\left(x\right)dx& =\underset{c}{\overset{a}{\int }}f\left(x\right)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx\\ & =0+-5\\ & =-5\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(A\right)-5$

20. Soal UTBK Tes Kompetensi Akademik SAINTEK 2019

Misalkan fungsi $f$ memenuhi $f(x+5)=f\left(x\right)$ untuk setiap $x\in R$. Jika $\underset{1}{\overset{5}{\int }}f\left(x\right)dx=3$ dan $\underset{-5}{\overset{-4}{\int }}f\left(x\right)dx=-2$ maka nilai $\underset{5}{\overset{15}{\int }}f\left(x\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& 10\\ \left(B\right)& 6\\ \left(C\right)& 5\\ \left(D\right)& 2\\ \left(E\right)& 1\end{array}$

Alternatif Pembahasan:

Show

Catatan calon guru yang mungkin bermanfaat tentang sifat integral tentu;

• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx+\underset{b}{\overset{c}{\int }}f\left(x\right)dx=\underset{a}{\overset{c}{\int }}f\left(x\right)dx$
• Jika $f$ periodik dengan periode $p$, maka $\underset{a+p}{\overset{b+p}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
  ${}^{\prime }$Suatu fungsi $f$ adalah periodik jika terdapat suatu bilangan $p$ sedemikian sehingga $f(x+p)=f\left(x\right)$${}^{\prime }$

Karena $f(x+5)=f\left(x\right)$ maka $f\left(x\right)$ periodik dengan periode $5$, sehingga berlaku:

• $\underset{1}{\overset{5}{\int }}f\left(x\right)dx=\underset{6}{\overset{10}{\int }}f\left(x\right)dx=\underset{11}{\overset{15}{\int }}f\left(x\right)dx=3$
• $\underset{-5}{\overset{-4}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{5}{\overset{6}{\int }}f\left(x\right)dx=\underset{10}{\overset{11}{\int }}f\left(x\right)dx=-2$;

Dengan menggunakan sifat integral di atas, maka berlaku
$\begin{array}{cc}\underset{5}{\overset{15}{\int }}f\left(x\right)dx& =\underset{5}{\overset{6}{\int }}f\left(x\right)dx+\underset{6}{\overset{10}{\int }}f\left(x\right)dx+\underset{10}{\overset{11}{\int }}f\left(x\right)dx+\underset{11}{\overset{15}{\int }}f\left(x\right)dx\\ & =-2+3+(-2)+3\\ & =2\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)2$

21. Soal UTBK Tes Kompetensi Akademik SAINTEK 2019

Diketahui fungsi $f\left(x\right)$ adalah fungsi genap, Jika nilai $\underset{-5}{\overset{5}{\int }}\left(f\left(x\right)+3x^2\right)dx=260$ dan $\underset{2}{\overset{4}{\int }}f\left(x\right)dx=2$ maka nilai $\underset{0}{\overset{2}{\int }}f\left(x\right)dx+\underset{4}{\overset{5}{\int }}f\left(x\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& -7\\ \left(B\right)& -3\\ \left(C\right)& 0\\ \left(D\right)& 3\\ \left(E\right)& 7\end{array}$

Alternatif Pembahasan:

Show

Catatan calon guru yang mungkin bermanfaat tentang sifat integral tentu;

• Berlaku $f(-x)=f\left(x\right)$
• Bentuk grafik fungsi, simetris dengan pusat sumbu $y$
• Jika dipakai pada integral, ciri fungsi genap ini adalah $\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

Pada soal disampaikan bahwa $f\left(x\right)$ adalah fungsi genap dan $3x^2$ adalah fungsi genap karena $f(-x)=f\left(x\right)$ sehingga berlaku:
$\begin{array}{cc}\underset{-5}{\overset{5}{\int }}\left(f\left(x\right)+3x^2\right)dx& =260\\ 2\cdot \underset{0}{\overset{5}{\int }}\left(f\left(x\right)+3x^2\right)dx& =260\\ \underset{0}{\overset{5}{\int }}\left(f\left(x\right)+3x^2\right)dx& =130\\ \underset{0}{\overset{5}{\int }}f\left(x\right)dx+\underset{0}{\overset{5}{\int }}3x^{2}dx& =130\\ \underset{0}{\overset{5}{\int }}f\left(x\right)dx+{\left|x^3\right|}_0^5& =130\\ \underset{0}{\overset{5}{\int }}f\left(x\right)dx+125& =130\\ \underset{0}{\overset{5}{\int }}f\left(x\right)& =5\end{array}$

$\begin{array}{cc}\underset{0}{\overset{5}{\int }}f\left(x\right)dx& =\underset{0}{\overset{2}{\int }}f\left(x\right)dx+\underset{2}{\overset{4}{\int }}f\left(x\right)dx+\underset{4}{\overset{5}{\int }}f\left(x\right)dx\\ 5& =\underset{0}{\overset{2}{\int }}f\left(x\right)dx+2+\underset{4}{\overset{5}{\int }}f\left(x\right)dx\\ 5-2& =\underset{0}{\overset{2}{\int }}f\left(x\right)dx+\underset{4}{\overset{5}{\int }}f\left(x\right)dx\\ 3& =\underset{0}{\overset{2}{\int }}f\left(x\right)dx+\underset{4}{\overset{5}{\int }}f\left(x\right)dx\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)3$

22. Soal UTBK Tes Kompetensi Akademik SAINTEK 2019

Diketahui $f(-x)=f\left(x\right)-3$ dan $x>0$. Jika $\underset{1}{\overset{5}{\int }}f\left(x\right)dx=2$ dan $\underset{3}{\overset{5}{\int }}f\left(x\right)dx=-3$ maka $\underset{-3}{\overset{-1}{\int }}f\left(x\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& -7\\ \left(B\right)& -1\\ \left(C\right)& 0\\ \left(D\right)& 1\\ \left(E\right)& 7\end{array}$

Alternatif Pembahasan:

Show

Catatan calon guru yang mungkin bermanfaat tentang integral tentu;

• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=-\underset{b}{\overset{a}{\int }}f\left(x\right)dx$
• $\underset{a}{\overset{b}{\int }}f(-x)dx=-\underset{-a}{\overset{-b}{\int }}f\left(x\right)dx$
• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx+\underset{b}{\overset{c}{\int }}f\left(x\right)dx=\underset{a}{\overset{c}{\int }}f\left(x\right)dx$

Diketahui $f(-x)=f\left(x\right)-3$ maka $f\left(x\right)=f(-x)+3$, sehingga berlaku:
$\begin{array}{cc}\underset{1}{\overset{5}{\int }}f\left(x\right)dx& =\underset{1}{\overset{5}{\int }}\left(f(-x)+3\right)dx\\ 2& =\underset{1}{\overset{5}{\int }}f(-x)dx+\underset{1}{\overset{5}{\int }}3dx\\ 2-\underset{1}{\overset{5}{\int }}3dx& =\underset{1}{\overset{5}{\int }}f(-x)dx\\ 2-{\left|3x\right|}_1^5& =-\underset{-1}{\overset{-5}{\int }}f\left(x\right)dx\\ 2-(15-3)& =-\underset{-1}{\overset{-5}{\int }}f\left(x\right)dx\\ -10& =\underset{-5}{\overset{-1}{\int }}f\left(x\right)dx\\ \underset{3}{\overset{5}{\int }}f\left(x\right)dx& =\underset{3}{\overset{5}{\int }}\left(f(-x)+3\right)dx\\ -3& =\underset{3}{\overset{5}{\int }}f(-x)dx+\underset{3}{\overset{5}{\int }}3dx\\ -3-\underset{3}{\overset{5}{\int }}3dx& =\underset{3}{\overset{5}{\int }}f(-x)dx\\ -3-{\left|3x\right|}_3^5& =-\underset{-3}{\overset{-5}{\int }}f\left(x\right)dx\\ -3-(15-9)& =-\underset{-3}{\overset{-5}{\int }}f\left(x\right)dx\\ -9& =-\underset{-3}{\overset{-5}{\int }}f\left(x\right)dx\\ 9& =\underset{-3}{\overset{-5}{\int }}f\left(x\right)dx\end{array}$

Dari persamaan yang kita peroleh di atas dan sifat integral tentu, dapat kita simpulkan:
$\begin{array}{cc}\underset{-3}{\overset{-1}{\int }}f\left(x\right)dx& =\underset{-3}{\overset{-5}{\int }}f\left(x\right)dx+\underset{-5}{\overset{-1}{\int }}f\left(x\right)dx\\ & =9-10\\ & =-1\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)-1$

23. Soal UTBK Tes Kompetensi Akademik SAINTEK 2019

Diberikan fungsi $f$ dengan $f(x+3)=f\left(x\right)$ untuk tiap $x$. Jika $\underset{-3}{\overset{6}{\int }}f\left(x\right)dx=-6$, maka $\underset{3}{\overset{9}{\int }}f\left(x\right)dx=\cdots$
$\begin{array}{cc}\left(A\right)& -4\\ \left(B\right)& -6\\ \left(C\right)& -8\\ \left(D\right)& -10\\ \left(E\right)& -12\end{array}$

Alternatif Pembahasan:

Show

Catatan calon guru yang mungkin bermanfaat tentang sifat integral tentu;

• $\underset{a}{\overset{b}{\int }}f\left(x\right)dx+\underset{b}{\overset{c}{\int }}f\left(x\right)dx=\underset{a}{\overset{c}{\int }}f\left(x\right)dx$
• Jika $f$ periodik dengan periode $p$, maka $\underset{a+p}{\overset{b+p}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
  ${}^{\prime }$Suatu fungsi $f$ adalah periodik jika terdapat suatu bilangan $p$ sedemikian sehingga $f(x+p)=f\left(x\right)$${}^{\prime }$

Karena $f(x+3)=f\left(x\right)$ maka $f\left(x\right)$ periodik dengan periode $3$, sehingga berlaku:
$\begin{array}{cc}\underset{3}{\overset{9}{\int }}f\left(x\right)dx& =\underset{0}{\overset{6}{\int }}f\left(x\right)\\ -6& =\underset{-3}{\overset{6}{\int }}f\left(x\right)dx\\ -6& =\underset{-3}{\overset{0}{\int }}f\left(x\right)dx+\underset{0}{\overset{3}{\int }}f\left(x\right)dx+\underset{3}{\overset{6}{\int }}f\left(x\right)dx\\ -6& =\underset{-3}{\overset{0}{\int }}f\left(x\right)dx+\underset{0}{\overset{3}{\int }}f\left(x\right)dx+\underset{3-3}{\overset{6-3}{\int }}f(x+3)dx\\ -6& =\underset{0}{\overset{3}{\int }}f\left(x\right)dx+\underset{0}{\overset{3}{\int }}f\left(x\right)dx+\underset{0}{\overset{3}{\int }}f\left(x\right)dx\\ -6& =3\underset{0}{\overset{3}{\int }}f\left(x\right)dx\\ -2& =\underset{0}{\overset{3}{\int }}f\left(x\right)dx\\ \underset{3}{\overset{9}{\int }}f\left(x\right)dx& =\underset{0}{\overset{6}{\int }}f\left(x\right)\\ & =\underset{0}{\overset{3}{\int }}f\left(x\right)+\underset{3}{\overset{6}{\int }}f\left(x\right)\\ & =\underset{0}{\overset{3}{\int }}f\left(x\right)+\underset{3-3}{\overset{6-3}{\int }}f(x+3)\\ & =\underset{0}{\overset{3}{\int }}f\left(x\right)+\underset{0}{\overset{3}{\int }}f\left(x\right)\\ & =(-2)+(-2)\\ & =-4\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(A\right)-4$

24. Soal SIMAK UI 2019 Kode 314/323 (*Soal Lengkap)

Jika $\underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)f\left(x\right)dx=10$ dan $f\left(a\right)=2+f\left(b\right)$, nilai $f\left(b\right)=\cdots$
$\begin{array}{cc}\left(A\right)& -2\\ \left(B\right)& -4\\ \left(C\right)& -6\\ \left(D\right)& -8\\ \left(E\right)& -10\end{array}$

Alternatif Pembahasan:

Show

Untuk menyelesaikan soal di atas, kita coba dengan memisalkan $u=f\left(x\right)$, sehingga beberapa persamaan yang kita peroleh, yaitu:
$\begin{array}{cc}u& =f\left(x\right)\\ \frac{du}{dx}& =f^{\prime }\left(x\right)\\ du& =f^{\prime }\left(x\right)dx\end{array}$
Dari pemisalan yang kita peroleh di atas;
$\begin{array}{cc}\underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)f\left(x\right)dx& =10\\ \underset{a}{\overset{b}{\int }}f\left(x\right)f^{\prime }\left(x\right)dx& =10\\ \underset{a}{\overset{b}{\int }}udu& =10\\ {\left[\frac{1}{2}\cdot u^2\right]}_a^b& =10\\ {\left[\frac{1}{2}\cdot {\left(f\left(x\right)\right)}^2\right]}_a^b& =10\\ \left[\frac{1}{2}\cdot {\left(f\left(b\right)\right)}^2\right]-\left[\frac{1}{2}\cdot {\left(f\left(a\right)\right)}^2\right]& =10\\ {\left(f\left(b\right)\right)}^2-{\left(f\left(a\right)\right)}^2& =20\\ \left(f\left(b\right)-f\left(a\right)\right)\left(f\left(b\right)+f\left(a\right)\right)& =20\\ \left(f\left(b\right)-2-f\left(b\right)\right)\left(f\left(b\right)+2+f\left(b\right)\right)& =20\\ \left(-2\right)\left(2f\left(b\right)+2\right)& =20\\ 2f\left(b\right)+2& =-10\\ 2f\left(b\right)& =-12\\ f\left(b\right)& =-6\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(C\right)-6$

25. Soal SBMPTN 2018 SAINTEK Kode 403 (*Soal Lengkap)

Jika $\underset{1}{\overset{2}{\int }}f\left(x\right)dx=\sqrt{2}$, maka nilai $\underset{1}{\overset{4}{\int }}\frac{1}{\sqrt{x}}f\left(\sqrt{x}\right)dx$ adalah...
$\begin{array}{cc}\left(A\right)& \frac{\sqrt{2}}{4}\\ \left(B\right)& \frac{\sqrt{2}}{2}\\ \left(C\right)& \sqrt{2}\\ \left(D\right)& 2\sqrt{2}\\ \left(E\right)& 4\sqrt{2}\end{array}$

Alternatif Pembahasan:

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$\begin{array}{cc}\underset{1}{\overset{2}{\int }}f\left(x\right)dx& =\sqrt{2}\\ {\left[F\left(x\right)\right]}_1^2& =\sqrt{2}\\ F\left(2\right)-F\left(1\right)& =\sqrt{2}\end{array}$

Selanjutnya, untuk menyelesaikan soal di atas, kita coba dengan memisalkan $u=\sqrt{x}$, sehingga beberapa persamaan kita peroleh, yaitu:
$\begin{array}{cc}u& =\sqrt{x}\\ \frac{du}{dx}& =\frac{1}{2\sqrt{x}}\\ du\cdot 2\sqrt{x}& =dx\\ du\cdot 2& =\frac{dx}{\sqrt{x}}\end{array}$
Dari pemisalan yang kita peroleh di atas;
$\begin{array}{cc}\underset{1}{\overset{4}{\int }}\frac{1}{\sqrt{x}}f\left(\sqrt{x}\right)dx& =\underset{1}{\overset{4}{\int }}f\left(\sqrt{x}\right)\frac{dx}{\sqrt{x}}\\ x=4\to u& =\sqrt{x}=\sqrt{4}=2\\ x=1\to u& =\sqrt{x}=\sqrt{1}=1\\ & =\underset{1}{\overset{2}{\int }}f\left(u\right)2du\\ & =2\underset{1}{\overset{2}{\int }}f\left(u\right)du\\ & =2\cdot {\left[F\left(u\right)\right]}_1^2\\ & =2\cdot \left(F\left(2\right)-F\left(1\right)\right)\\ & =2\cdot \left(\sqrt{2}\right)\\ & =2\sqrt{2}\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(D\right)2\sqrt{2}$

$\begin{array}{cc}\underset{2}{\overset{3}{\int }}f\left(x\right)dx& =\sqrt{2}\\ {\left[F\left(x\right)\right]}_2^3& =\sqrt{2}\\ F\left(3\right)-F\left(2\right)& =\sqrt{2}\end{array}$

Selanjutnya, untuk menyelesaikan soal di atas, kita coba dengan memisalkan $u=1+\frac{2}{x}$, sehingga beberapa persamaan kita peroleh, yaitu:
$\begin{array}{cc}u& =1+\frac{2}{x}\\ \frac{du}{dx}& =-\frac{2}{x^2}\\ -\frac{du}{2}& =\frac{2}{x^2}\end{array}$

Dari pemisalan yang kita peroleh di atas;
$\begin{array}{cc}\underset{1}{\overset{2}{\int }}\frac{1}{x^2}f\left(1+\frac{2}{x}\right)dx& =\underset{1}{\overset{2}{\int }}f\left(1+\frac{2}{x}\right)\frac{dx}{x^2}\\ x=2\to u& =1+\frac{2}{x}=1+\frac{2}{2}=2\\ x=1\to u& =1+\frac{2}{x}=1+\frac{2}{1}=3\\ & =\underset{3}{\overset{2}{\int }}f\left(u\right)\cdot -\frac{du}{2}\\ & =-\frac{1}{2}\underset{3}{\overset{2}{\int }}f\left(u\right)du\\ & =-\frac{1}{2}{\left[F\left(u\right)\right]}_3^2\\ & =-\frac{1}{2}\left(F\left(2\right)-F\left(3\right)\right)\\ & =\frac{1}{2}\left(F\left(3\right)-F\left(2\right)\right)\\ & =\frac{1}{2}\left(\sqrt{2}\right)\end{array}$

$\therefore$ Pilihan yang sesuai adalah $\left(B\right)\frac{\sqrt{2}}{2}$