HOTS Zone : Integral Tentu

Berikut ini adalah kumpulan soal mengenai integral tentu tipe HOTS. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

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No. 1

\displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2016x}\ dx= ....

Brilliant.org

Jadi, \displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2016x}\ dx=\sqrt2.

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No. 2

Misalkan fungsi f memenuhi
x=f(x)e^{f(x)}
Nilai dari
\displaystyle\intop_0^ef(x)\ dx=

Untuk x=0,
$\begin{array}{rl}0& =f(0)e^{f(0)}\\ f(0)& =0\end{array}$

Untuk x=e,
$\begin{array}{rl}e& =f(e)e^{f(e)}\\ f(e)& =1\end{array}$

$\begin{array}{rl}{\displaystyle \underset{0}{\overset{e}{\int }}f(x)dx}& ={\displaystyle \underset{0}{\overset{e}{\int }}f(x)d(f(x)e^{f(x)})}\\ & ={\displaystyle \underset{0}{\overset{e}{\int }}f(x)(e^{f(x)}d(f(x))+f(x)e^{f(x)}d(f(x)))}\\ & ={\displaystyle \underset{0}{\overset{e}{\int }}(f(x)e^{f(x)}+f^2(x)e^{f(x)})d(f(x))}\\ & ={\displaystyle \underset{0}{\overset{e}{\int }}(u{e}^u+u^2{e}^u)du}\\ & ={\left[{\displaystyle \underset{0}{\overset{e}{\int }}u{e}^{u}du+u^2{e}^u-{\displaystyle \underset{0}{\overset{e}{\int }}2u{e}^{u}du}}\right]}_0^e\\ & ={[u^2{e}^u-{\displaystyle \underset{0}{\overset{e}{\int }}u{e}^{u}du}]}_0^e\\ & ={[u^2{e}^u-u{e}^u+e^u]}_0^e\\ & ={[f^2(x)e^{f(x)}-f(x)e^{f(x)}+e^{f(x)}]}_0^e\\ & =[f^2(e)e^{f(e)}-f(e)e^{f(e)}+e^{f(e)}]-[f^2(0)e^{f(0)}-f(0)e^{f(0)}+e^{f(0)}]\\ & =[(1{)}^2{e}^1-(1)e^1+e^1]-[(0{)}^2{e}^0-(0)e^0+e^0]\\ & =[e-e+e]-[0-0+1]\\ & =\boxed{{\displaystyle \boxed{{\displaystyle e-1}}}}\end{array}$

Jadi, \displaystyle\intop_0^ef(x)\ dx=e-1.