HOTS Zone : Integral Tentu
Table of Contents
Tipe:
No.
\[\displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2020x}\ dx=....\]ALTERNATIF PENYELESAIAN
Misal 2020x = 4kx = t
dt = 4k dx
\(\begin{aligned} \displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2016x}\ dx&=\displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin4kx}\ dx\\ &=\dfrac1{4k}\intop_0^{2k\pi}\sqrt{1-\sin t}\ dt\\ &=\frac{k}{4k}\intop_0^{2\pi}\sqrt{1-\sin t}\ dt&{\color{red}\text{karena }\sqrt{1-\sin t}\text{ mempunyai periode }2\pi}\\ &=\frac14\left(\intop_0^{\pi}\sqrt{1-\sin t}\ dt+\intop_{\pi}^{2\pi}\sqrt{1-\sin t}\ dt\right)\\ &=\frac14\left(2\intop_0^{\frac{\pi}2}\sqrt{1-\sin t}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\sin t}\ dt\right)\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt{1-\sin\left(0+\frac{\pi}2-x\right)}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\sin\left(\pi+\frac{3\pi}2-x\right)}\ dt\right)&{\color{red}\intop_a^b f(x)\ dx=\intop_a^b f(a+b-x)\ dx}\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt{1-\cos t}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\cos t}\ dt\right)&{\color{red}\sin\left(\frac{\pi}2-\theta\right)=\cos \theta}\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt2\sin\frac{t}2\ dt+\intop_{\pi}^{\frac{3\pi}2}\sqrt2\sin\frac{t}2\ dt\right)&{\color{red}\cos2\theta=1-\sin^2\theta}\\ &=\sqrt2\left(\left.\cos\frac{t}2\right|^0_{\frac{\pi}2}+\left.\cos\frac{t}2\right|^{\pi}_{\frac{3\pi}2}\right)&{\color{red}\int\sin\frac{\theta}2\ d\theta=-2\cos\frac{\theta}2+C}\\ &=\sqrt2\left(1-\frac1{\sqrt2}+0+\frac1{\sqrt2}\right)\\ &=\boxed{\boxed{\sqrt2}} \end{aligned}\)
dt = 4k dx
\(\begin{aligned} \displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2016x}\ dx&=\displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin4kx}\ dx\\ &=\dfrac1{4k}\intop_0^{2k\pi}\sqrt{1-\sin t}\ dt\\ &=\frac{k}{4k}\intop_0^{2\pi}\sqrt{1-\sin t}\ dt&{\color{red}\text{karena }\sqrt{1-\sin t}\text{ mempunyai periode }2\pi}\\ &=\frac14\left(\intop_0^{\pi}\sqrt{1-\sin t}\ dt+\intop_{\pi}^{2\pi}\sqrt{1-\sin t}\ dt\right)\\ &=\frac14\left(2\intop_0^{\frac{\pi}2}\sqrt{1-\sin t}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\sin t}\ dt\right)\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt{1-\sin\left(0+\frac{\pi}2-x\right)}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\sin\left(\pi+\frac{3\pi}2-x\right)}\ dt\right)&{\color{red}\intop_a^b f(x)\ dx=\intop_a^b f(a+b-x)\ dx}\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt{1-\cos t}\ dt+2\intop_{\pi}^{\frac{3\pi}2}\sqrt{1-\cos t}\ dt\right)&{\color{red}\sin\left(\frac{\pi}2-\theta\right)=\cos \theta}\\ &=\frac12\left(\intop_0^{\frac{\pi}2}\sqrt2\sin\frac{t}2\ dt+\intop_{\pi}^{\frac{3\pi}2}\sqrt2\sin\frac{t}2\ dt\right)&{\color{red}\cos2\theta=1-\sin^2\theta}\\ &=\sqrt2\left(\left.\cos\frac{t}2\right|^0_{\frac{\pi}2}+\left.\cos\frac{t}2\right|^{\pi}_{\frac{3\pi}2}\right)&{\color{red}\int\sin\frac{\theta}2\ d\theta=-2\cos\frac{\theta}2+C}\\ &=\sqrt2\left(1-\frac1{\sqrt2}+0+\frac1{\sqrt2}\right)\\ &=\boxed{\boxed{\sqrt2}} \end{aligned}\)
Jadi, \displaystyle\intop_0^{\frac{\pi}2}\sqrt{1-\sin2020x}\ dx=\sqrt2 .
No.
Misalkan fungsi f memenuhiALTERNATIF PENYELESAIAN
Untuk x = 0,
\(\begin{aligned} 0&=f(0)e^{f(0)}\\ f(0)&=0 \end{aligned}\)
Untuk x = e,
\(\begin{aligned} e&=f(e)e^{f(e)}\\ f(e)&=1 \end{aligned}\)
\(\begin{aligned} \displaystyle\intop_0^ef(x)\ dx&=\displaystyle\intop_0^ef(x)\ d\left(f(x)e^{f(x)}\right)\\ &=\displaystyle\intop_0^ef(x)\left(e^{f(x)}\ d\left(f(x)\right)+f(x)e^{f(x)}\ d\left(f(x)\right)\right)\\ &=\displaystyle\intop_0^e\left(f(x)e^{f(x)}+f^2(x)e^{f(x)}\right)\ d\left(f(x)\right)\\ &=\displaystyle\intop_0^e\left(ue^u+u^2e^u\right)\ du\\ &=\left[\displaystyle\intop_0^eue^u\ du+u^2e^u-\displaystyle\intop_0^e2ue^u\ du\right]_0^e\\ &=\left[u^2e^u-\displaystyle\intop_0^eue^u\ du\right]_0^e\\ &=\left[u^2e^u-ue^u+e^u\right]_0^e\\ &=\left[f^2(x)e^{f(x)}-f(x)e^{f(x)}+e^{f(x)}\right]_0^e\\ &=\left[f^2(e)e^{f(e)}-f(e)e^{f(e)}+e^{f(e)}\right]-\left[f^2(0)e^{f(0)}-f(0)e^{f(0)}+e^{f(0)}\right]\\ &=\left[(1)^2e^1-(1)e^1+e^1\right]-\left[(0)^2e^0-(0)e^0+e^0\right]\\ &=\left[e-e+e\right]-\left[0-0+1\right]\\ &=\boxed{\boxed{e-1}} \end{aligned}\)
\(\begin{aligned} 0&=f(0)e^{f(0)}\\ f(0)&=0 \end{aligned}\)
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 .
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