Berikut ini adalah pembahasan soal integral SBMPTN 2018 TKD SAINTEK.
1. $($ SBMPTN 2018 Kode 453 $)$ Jika $\int_{1}^{2}f(x)dx=\sqrt{2}$, maka nilai $\int_{1}^{4}\frac{1}{\sqrt{x}}f(\sqrt{x})dx$ adalah ...
A. $\frac{\sqrt{2}}{4}$
B. $\frac{\sqrt{2}}{2}$
C. $\sqrt{2}$
D. $2\sqrt{2}$
E. $4\sqrt{2}$

Jawab D
Pembahasan :
Misal $u=\sqrt{x}$
$du=\frac{1}{2\sqrt{x}}dx$.

Merubah batas integral
$x=1 \rightarrow u=1$
$x=4 \rightarrow u=2$
sehingga diperoleh
\begin{align}
\int_{1}^{4}\frac{1}{\sqrt{x}}f(\sqrt{x})dx &=\int_{1}^{2}\frac{1}{\sqrt{x}}f(u)\frac{du}{\frac{1}{2\sqrt{x}}}\\
&=\int_{1}^{2}2f(u)du\\
&=2\sqrt{2}.
\end{align}

2. $($ SBMPTN 2018 Kode 454 $)$ Jika $\int_{2}^{3}f(x)dx=\sqrt{2}$, maka nilai $\int_{1}^{2}\frac{1}{x^{2}}f(1+\frac{2}{x})dx$ adalah ...
A. $\frac{\sqrt{2}}{4}$
B. $\frac{\sqrt{2}}{2}$
C. $\sqrt{2}$
D. $2\sqrt{2}$
E. $4\sqrt{2}$

Jawab B
Pembahasan :
Misal $u=1+\frac{2}{x}$
$du=\frac{-2}{x^{2}}dx$.

Merubah batas integral
$x=1 \rightarrow u=3$
$x=2 \rightarrow u=2$
sehingga diperoleh
\begin{align}
\int_{1}^{2}\frac{1}{x^{2}}f(1+\frac{2}{x})dx&=\int_{3}^{2}-\frac{1}{2}f(u)du\\
&=\int_{2}^{3}\frac{1}{2}f(u)du\\
&=\frac{1}{2}\int_{2}^{3}f(u)du\\
&=\frac{\sqrt{2}}{2}.
\end{align}

3. $($ SBMPTN 2018 Kode 455 $)$ Nilai $\int_{1/8}^{1/3}\frac{3}{x^{2}}\sqrt{1+\frac{1}{x}}$ adalah ...
A. 19
B. 38
C. 57
D. 76
E. 95

Jawab B
Pembahasan :
Misal $u=1+\frac{1}{x}$
$du=-\frac{1}{x^{2}}dx$.

Merubah batas integral
$x=1/8 \rightarrow u=9$
$x=1/3 \rightarrow u=4$

sehingga diperoleh
\begin{align}
\int_{1/8}^{1/3}\frac{3}{x^{2}}\sqrt{1+\frac{1}{x}}&=\int_{9}^{4}\frac{3}{x^{2}}\sqrt{u}\frac{du}{-\frac{1}{x^{2}}}\\
&=\int_{9}^{4}-3\sqrt{u}du\\
&=3\int_{4}^{9}\sqrt{u}du\\
&=3\int_{4}^{9} u^{\frac{1}{2}}du\\
&=(2u^{\frac{3}{2}})_{4}^{9}\\
&=2(27-8)\\
&=38.
\end{align}