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• Notes

Calculate the iterated integral.
$\int_{1}^{4} \int_{0}^{2}\left(6 x^{2} y-2 x\right) d y d x$

$\int_{1}^{4} \int_{0}^{2}\left(6 x^{2} y-2 x\right) d y d x=\int_{0}^{2} \int_{1}^{4}\left(6 x^{2} y-2 x\right) d x d y$

$=\int_{1}^{4} |_{0}^{4} 6 x \frac{y^2}{2}-2 x y d x$

$=\int_{1}^{4} 6 x^{2} \frac{(2)^{2}}{2}-2 * 2 x d x$

$=\int_{1}^{4} \left(12 x^{2}-4 x\right) d x$

$|_{1}^{4} \frac{12}{3} x^{3}-4 \frac{x^{2}}{2}=222$

$\int_{0}^{2} \int_{1}^{4}\left(6 x^{2} y-2 x\right) d x d y$

$=\int_{0}^{2} |_{1}^{4} \frac{6 x^{3}}{3}y-2 \frac{x^{2}}{2} d y$

$=\int_{0}^{2} \left.=2\left[4^{3}-1^{3}\right]y-\left[4^{2}-(1\right)^{2}\right] d y$

$=\int_{0}^{2} 126 y-15 d y=|_{0}^{2} \frac {126}{2} y^{2}-15 y=222$

$\iint_{R} x \sec ^{2} y d A, \quad R=\{(x, y) | 0 \leq x \leq 2,0 \leq y \leq \pi / 4\}$

$\int_{0}^{2} \int_{0}^{\pi / 4} x \sec ^{2} y d x d y$

$=\int_{0}^{2} x d x \int_{0}^{\pi /4} \sec ^{2} y d y$

$=\left|_{0}^{2}\frac{x^{2}}{2} *\right|_{0}^{\pi / 4} \tan y=\frac{(2)^{2}}{2} * \tan \left(\frac{\pi}{4}\right)=2$