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. 1

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

Grup Telegram

Misal:

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$$\begin{array}{rl}u& =x\\ du& =dx\end{array}$$

$$\begin{array}{rl}dv& =\sin (xy)dx\\ v& =-{\displaystyle \frac{1}{y}}\cos (xy)\end{array}$$

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$$\begin{array}{rl}{\displaystyle \int udv}& =uv-{\displaystyle \int vdu}\\ {\displaystyle \int x\sin (xy)dx}& =x(-{\displaystyle \frac{1}{y}}\cos (xy))-{\displaystyle \int (-{\displaystyle \frac{1}{y}}\cos (xy))dx}\\ & =-{\displaystyle \frac{x\cos (xy)}{y}}+{\displaystyle \frac{1}{y}}{\displaystyle \int \cos (xy)dx}\\ & =-{\displaystyle \frac{x\cos (xy)}{y}}+{\displaystyle \frac{\sin (xy)}{y^2}}+C\\ & ={\displaystyle \frac{1}{y^2}}(-xy\cos (xy)+\sin (xy))+C\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{4}{\int }}\underset{\pi }{\overset{\frac{3\pi }{2}}{\int }}x\sin (xy)dxdy}& ={\displaystyle \underset{0}{\overset{4}{\int }}{\displaystyle \frac{1}{y^2}}{[-xy\cos (xy)+\sin (xy)]}_{\pi }^{\frac{3\pi }{2}}dy}\\ & ={\displaystyle \underset{0}{\overset{4}{\int }}{\displaystyle \frac{1}{y^2}}[(-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right))-(-\pi y\cos (\pi y)+\sin \left(\pi y\right))]dy}\\ & ={\displaystyle \underset{0}{\overset{4}{\int }}(-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right)+\pi y\cos (\pi y)-\sin \left(\pi y\right)){\displaystyle \frac{1}{y^2}}dy}\\ & ={\displaystyle \underset{0}{\overset{4}{\int }}((-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right)){\displaystyle \frac{1}{y^2}}+(\pi y\cos (\pi y)-\sin \left(\pi y\right)){\displaystyle \frac{1}{y^2}})dy}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \int (-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right)){\displaystyle \frac{1}{y^2}}dy}& =(-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right))(-{\displaystyle \frac{1}{y}})-{\displaystyle \int (-{\displaystyle \frac{1}{y}})({\displaystyle \frac{9{\pi }^2}{4}}y\sin \left({\displaystyle \frac{3\pi }{2}}y\right))dy}\\ & ={\displaystyle \frac{3\pi }{2}}\cos \left({\displaystyle \frac{3\pi }{2}}y\right)-{\displaystyle \frac{\sin \left({\displaystyle \frac{3\pi }{2}}y\right)}{y}}+{\displaystyle \frac{9{\pi }^2}{4}}{\displaystyle \int \sin \left({\displaystyle \frac{3\pi }{2}}y\right)dy}\\ & ={\displaystyle \frac{3\pi }{2}}\cos \left({\displaystyle \frac{3\pi }{2}}y\right)-{\displaystyle \frac{\sin \left({\displaystyle \frac{3\pi }{2}}y\right)}{y}}+{\displaystyle \frac{9{\pi }^2}{4}}(-{\displaystyle \frac{2}{3\pi }}\cos \left({\displaystyle \frac{3\pi }{2}}y\right))+C\\ & ={\displaystyle \frac{3\pi }{2}}\cos \left({\displaystyle \frac{3\pi }{2}}y\right)-{\displaystyle \frac{\sin \left({\displaystyle \frac{3\pi }{2}}y\right)}{y}}-{\displaystyle \frac{3\pi }{2}}\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+C\\ & =-{\displaystyle \frac{\sin \left({\displaystyle \frac{3\pi }{2}}y\right)}{y}}+C\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{4}{\int }}((-{\displaystyle \frac{3\pi }{2}}y\cos \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \left({\displaystyle \frac{3\pi }{2}}y\right)){\displaystyle \frac{1}{y^2}}+(\pi y\cos (\pi y)-\sin \left(\pi y\right)){\displaystyle \frac{1}{y^2}})dy}& ={\left[{\displaystyle \frac{-\sin \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \pi y}{y}}\right]}_0^4\\ & ={\displaystyle \frac{-\sin ({\displaystyle \frac{3\pi }{2}}(4))+\sin \pi (4)}{4}}-{\displaystyle \underset{y\to 0}{lim}\left({\displaystyle \frac{-\sin \left({\displaystyle \frac{3\pi }{2}}y\right)+\sin \pi y}{y}}\right)}\\ & ={\displaystyle \frac{-\sin 6\pi +\sin 4\pi }{4}}-(-{\displaystyle \frac{3\pi }{2}}+\pi )\\ & ={\displaystyle \frac{-0+0}{4}}-(-{\displaystyle \frac{\pi }{2}})\\ & =0+{\displaystyle \frac{\pi }{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\pi }{2}}}}}}\end{array}$$

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