Question

1) (23-36-) + (1) + (8426

Answer 1

Compute the integral with respect to x, then with respect to y:

$\displaystyle16\int_0^\pi\int_0^1x^2\sin y\,\mathrm dx\,\mathrm dy=16\int_0^\pi\sin y\frac{x^3}3\bigg|_0^1\,\mathrm dy$

$=\displaystyle\frac{16}3\int_0^\pi\sin y\,\mathrm dy$

$=\displaystyle\frac{16}3(-\cos y)\bigg|_0^\pi=\boxed{\dfrac{32}3}$

Alternatively, in this case you can "factorize" the integral as

$\displaystyle16\left(\int_0^\pi\sin y\,\mathrm dy\right)\left(\int_0^1x^2\,\mathrm dx\right)$

and get the same result.