Exercise Zone : Integral Lipat

Berikut ini adalah kumpulan soal mengenai Integral Lipat. Jika ingin bertanya soal, silahkan gabung ke grup Telegram.

Tipe:

Standar SBMPTN HOTS

No.

\displaystyle\intop_0^4\intop_{\pi}^{\frac{3\pi}2}x\sin(xy)\ dx\ dy

Grup Telegram

Misal:

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\begin{aligned} u&=x\\ du&=dx \end{aligned}

\begin{aligned} dv&=\sin(xy)\ dx\\ v&=-\dfrac1y\cos(xy) \end{aligned}

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\begin{aligned} \displaystyle\int u\ dv&=uv-\displaystyle\int v\ du\\ \displaystyle\int x\sin(xy)\ dx&=x\left(-\dfrac1y\cos(xy)\right)-\displaystyle\int\left(-\dfrac1y\cos(xy)\right)\ dx\\ &=-\dfrac{x\cos(xy)}y+\dfrac1y\displaystyle\int\cos(xy)\ dx\\ &=-\dfrac{x\cos(xy)}y+\dfrac{\sin(xy)}{y^2}+C\\ &=\dfrac1{y^2}\left(-xy\cos(xy)+\sin(xy)\right)+C \end{aligned}

\begin{aligned} \displaystyle\intop_0^4\intop_{\pi}^{\frac{3\pi}2}x\sin(xy)\ dx\ dy&=\displaystyle\intop_0^4\dfrac1{y^2}\left[-xy\cos(xy)+\sin(xy)\right]_{\pi}^{\frac{3\pi}2}\ dy\\ &=\displaystyle\intop_0^4\dfrac1{y^2}\left[\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)\right)-\left(-\pi y\cos(\pi y)+\sin\left(\pi y\right)\right)\right]\ dy\\ &=\displaystyle\intop_0^4\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)+\pi y\cos(\pi y)-\sin\left(\pi y\right)\right)\dfrac1{y^2}\ dy\\ &=\displaystyle\intop_0^4\left(\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)\right)\dfrac1{y^2}+\left(\pi y\cos(\pi y)-\sin\left(\pi y\right)\right)\dfrac1{y^2}\right)\ dy\\ \end{aligned}

u	dv
-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)	\dfrac1{y^2}=y^{-2}
-\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)+\left(-\dfrac{3\pi}2y\right)\left(-\dfrac{3\pi}2\sin\left(\dfrac{3\pi}2y\right)\right)+\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)=\dfrac{9\pi^2}4y\sin\left(\dfrac{3\pi}2y\right)	\dfrac1{-1}y^{-1}=-\dfrac1y

\begin{aligned} \displaystyle\int\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)\right)\dfrac1{y^2}\ dy&=\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)\right)\left(-\dfrac1y\right)-\displaystyle\int\left(-\dfrac1y\right)\left(\dfrac{9\pi^2}4y\sin\left(\dfrac{3\pi}2y\right)\right)\ dy\\ &=\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)-\dfrac{\sin\left(\dfrac{3\pi}2y\right)}y+\dfrac{9\pi^2}4\displaystyle\int\sin\left(\dfrac{3\pi}2y\right)\ dy\\ &=\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)-\dfrac{\sin\left(\dfrac{3\pi}2y\right)}y+\dfrac{9\pi^2}4\left(-\dfrac2{3\pi}\cos\left(\dfrac{3\pi}2y\right)\right)+C\\ &=\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)-\dfrac{\sin\left(\dfrac{3\pi}2y\right)}y-\dfrac{3\pi}2\cos\left(\dfrac{3\pi}2y\right)+C\\ &=-\dfrac{\sin\left(\dfrac{3\pi}2y\right)}y+C \end{aligned}

{\displaystyle\int\left(\pi y\cos \pi y+\sin \pi y\right)\dfrac1{y^2}\ dy=\dfrac{\sin \pi y}y+C}

\begin{aligned} \displaystyle\intop_0^4\left(\left(-\dfrac{3\pi}2y\cos\left(\dfrac{3\pi}2y\right)+\sin\left(\dfrac{3\pi}2y\right)\right)\dfrac1{y^2}+\left(\pi y\cos(\pi y)-\sin\left(\pi y\right)\right)\dfrac1{y^2}\right)\ dy&=\left[\dfrac{-\sin\left(\dfrac{3\pi}2y\right)+\sin \pi y}y\right]_0^4\\ &=\dfrac{-\sin\left(\dfrac{3\pi}2(4)\right)+\sin \pi (4)}4-\displaystyle\lim_{y\to0}\left(\dfrac{-\sin\left(\dfrac{3\pi}2y\right)+\sin \pi y}y\right)\\ &=\dfrac{-\sin6\pi+\sin 4\pi}4-\left(-\dfrac{3\pi}2+\pi\right)\\ &=\dfrac{-0+0}4-\left(-\dfrac{\pi}2\right)\\ &=0+\dfrac{\pi}2\\ &=\boxed{\boxed{\dfrac{\pi}2}} \end{aligned}

Jadi, \displaystyle\intop_0^4\intop_{\pi}^{\frac{3\pi}2}x\sin(xy)\ dx\ dy=\dfrac{\pi}2.