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

---

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.

---

No. 2

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

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

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

\(\begin{aligned} \displaystyle\intop_0^ef(x)\ dx&=\displaystyle\intop_0^ef(x)\ d\left(f(x)e^{f(x)}\right)\\ &=\displaystyle\intop_0^ef(x)\left(e^{f(x)}\ d\left(f(x)\right)+f(x)e^{f(x)}\ d\left(f(x)\right)\right)\\ &=\displaystyle\intop_0^e\left(f(x)e^{f(x)}+f^2(x)e^{f(x)}\right)\ d\left(f(x)\right)\\ &=\displaystyle\intop_0^e\left(ue^u+u^2e^u\right)\ du\\ &=\left[\displaystyle\intop_0^eue^u\ du+u^2e^u-\displaystyle\intop_0^e2ue^u\ du\right]_0^e\\ &=\left[u^2e^u-\displaystyle\intop_0^eue^u\ du\right]_0^e\\ &=\left[u^2e^u-ue^u+e^u\right]_0^e\\ &=\left[f^2(x)e^{f(x)}-f(x)e^{f(x)}+e^{f(x)}\right]_0^e\\ &=\left[f^2(e)e^{f(e)}-f(e)e^{f(e)}+e^{f(e)}\right]-\left[f^2(0)e^{f(0)}-f(0)e^{f(0)}+e^{f(0)}\right]\\ &=\left[(1)^2e^1-(1)e^1+e^1\right]-\left[(0)^2e^0-(0)e^0+e^0\right]\\ &=\left[e-e+e\right]-\left[0-0+1\right]\\ &=\boxed{\boxed{e-1}} \end{aligned}\)

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