[tex]\begin{aligned}\sf1.\ &\int_1^2\int_0^3xy\,dxdy=\bf\frac{27}{4}\\\sf2.\ &\int_1^2\int_0^3\left(xy+y^2\right)dxdy=\bf\frac{55}{4}\\\sf3.\ &\int_0^1\int_0^{3x}x^2\,dydx=\bf\frac{3}{4}\\\sf4.\ &\int_0^1\int_0^{3x}\left(x^2+x^2y^3\right)dydx=\bf\frac{51}{14}\end{aligned}[/tex]
Pembahasan
Integral Lipat/Ganda
Melihat bentuk integralnya, kita akan menggunakan hasil nomor 1 pada penyelesaian nomor 2, dan hasil nomor 3 pada penyelesaian nomor 4.
Nomor 1
[tex]\begin{aligned}&\int_1^2\int_0^3xy\,dxdy\\&{=\ }\int_1^2\left(\int_0^3xydx\right)dy\\&{=\ }\int_1^2\left[\frac{x^2y}{2}\right]_0^3dy\\&{=\ }\int_1^2\left[\frac{3^2y}{2}-\frac{0^2y}{2}\right]dy\\&{=\ }\int_1^2\frac{9y}{2}dy\\&{=\ }\frac{9}{2}\int_1^2y\,dy\\&{=\ }\frac{9}{2}\left[\frac{y^2}{2}\right]_1^2\\&{=\ }\frac{9}{2}\left[\frac{2^2}{2}-\frac{1^2}{2}\right]\\&{=\ }\frac{9}{2}\cdot\frac{3}{2}\\&{=\ }\boxed{\ \bf\frac{27}{4}\ }\end{aligned}[/tex]
Nomor 2
[tex]\begin{aligned}&\int_1^2\int_0^3\left(xy+y^2\right)dxdy\\&...\ \textsf{sifat penjumlahan pada integral}\\&{=\ }\underbrace{\int_1^2\int_0^3xy\,dxdy}_{\begin{array}{c}\textsf{dari nomor 1}\end{array}}+\int_1^2\int_0^3y^2\,dxdy\\&{=\ }\frac{27}{4}+\int_1^2\left(\int_0^3y^2\,dx\right)dy\\&{=\ }\frac{27}{4}+\int_1^2\left[xy^2\right]_0^3dy\\&{=\ }\frac{27}{4}+\int_1^2\left[3y^2-0y^2\right]dy\\&{=\ }\frac{27}{4}+\int_1^23y^2\,dy\\&{=\ }\frac{27}{4}+3\int_1^2y^2\,dy\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{27}{4}+3\left[\frac{y^3}{3}\right]_1^2\\&{=\ }\frac{27}{4}+3\left[\frac{2^3}{3}-\frac{1^3}{3}\right]\\&{=\ }\frac{27}{4}+3\left[\frac{7}{3}\right]\\&{=\ }\frac{27}{4}+7\\&{=\ }\boxed{\ \bf\frac{55}{4}\ }\end{aligned}[/tex]
Nomor 3
[tex]\begin{aligned}&\int_0^1\int_0^{3x}x^2\,dydx\\&{=\ }\int_0^1\left(\int_0^{3x}x^2dy\right)dx\\&{=\ }\int_0^1\left[yx^2\right]_0^{3x}dx\\&{=\ }\int_0^1\left[3x\cdot x^2-0x^2\right]dx\\&{=\ }\int_0^13x^3dx\\&{=\ }3\int_0^1x^3dx\\&{=\ }3\left[\frac{x^4}{4}\right]_0^1\\&{=\ }3\left[\frac{1^4}{4}-\frac{0^4}{4}\right]\\&{=\ }3\cdot\frac{1}{4}\\&{=\ }\boxed{\ \bf\frac{3}{4}\ }\end{aligned}[/tex]
Nomor 4
[tex]\begin{aligned}&\int_0^1\int_0^{3x}\left(x^2+x^2y^3\right)dydx\\&{=\ }\underbrace{\int_0^1\int_0^{3x}x^2\,dydx}_{\begin{array}{c}\textsf{dari nomor 3}\end{array}}+\int_0^1\int_0^{3x}x^2y^3\,dydx\\&{=\ }\frac{3}{4}+\int_0^1\left(\int_0^{3x}x^2y^3dy\right)dx\\&{=\ }\frac{3}{4}+\int_0^1\left[\frac{x^2y^4}{4}\right]_0^{3x}dx\\&{=\ }\frac{3}{4}+\int_0^1\left[\frac{x^2(3x)^4}{4}-\frac{x^2\left(0^4\right)}{4}\right]dx\\&{=\ }\frac{3}{4}+\int_0^1\frac{81x^6}{4}dx\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{3}{4}+\frac{81}{4}\int_0^1x^6dx\\&{=\ }\frac{3}{4}+\frac{81}{4}\left[\frac{x^7}{7}\right]_0^1\\&{=\ }\frac{3}{4}+\frac{81}{4}\left[\frac{1^7}{7}-\frac{0^7}{7}\right]\\&{=\ }\frac{3}{4}+\frac{81}{4}\cdot\frac{1}{7}\\&{=\ }\frac{3}{4}+\frac{81}{28}\\&{=\ }\frac{102}{28}\\&{=\ }\boxed{\ \bf\frac{51}{14}\ }\end{aligned}[/tex]
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