# How about solution. ( I = ?)

## Apr 28, 2018

$I = - {10}^{4} {e}^{- 2} \sin \left(2\right)$

#### Explanation:

We wish to know what the following integral evaluates to:

$I = {\int}_{0}^{2} {\int}_{0}^{1} - {10}^{4} \cos \left(2 x\right) {e}^{- 2} \mathrm{dx} \mathrm{dz}$

Start by removing the constants from the integrand.

$I = - {10}^{4} {e}^{- 2} {\int}_{0}^{2} {\int}_{0}^{1} \cos \left(2 x\right) \mathrm{dx} \mathrm{dz}$

We will let $C = - {10}^{4} {e}^{- 2}$ so that our integral is visually easier to work with.

Integrate $\cos \left(2 x\right)$ with respect to $x$ and evaluate from 0 to 1.

$I = C {\int}_{0}^{2} {\left[\frac{1}{2} \sin \left(2 x\right)\right]}_{0}^{1} \mathrm{dz} = \frac{C}{2} {\int}_{0}^{2} \sin \left(2\right) \mathrm{dz}$

Integrate the (constant) term $\sin \left(2\right)$ with respect to $z$ and evaluate from 0 to 2.

$I = \frac{C}{2} {\left[\sin \left(2\right) z\right]}_{0}^{2} = \frac{C}{2} 2 \sin \left(2\right) = C \sin \left(2\right)$

Finally, substitute our original constants back into $C$.

$I = - {10}^{4} {e}^{- 2} \sin \left(2\right)$

Apr 28, 2018

Given

$I = {\int}_{0}^{2} {\int}_{0}^{1} - {10}^{4} \cos \left(2 x\right) {e}^{- 2} \mathrm{dx} \mathrm{dz}$
$\implies I = - {10}^{4} {e}^{- 2} {\int}_{0}^{2} {\int}_{0}^{1} \cos \left(2 x\right) \mathrm{dx} \mathrm{dz}$

First Integrate outer integral with respect to $z$

$I = - {10}^{4} {e}^{- 2} | \left({\int}_{0}^{1} \cos \left(2 x\right) \mathrm{dx}\right) z {|}_{0}^{2}$
$\implies I = - {10}^{4} {e}^{- 2} \times 2 {\int}_{0}^{1} \cos \left(2 x\right) \mathrm{dx}$

Now Integrate with respect to $x$

$I = - {10}^{4} {e}^{- 2} \times 2 | \frac{1}{2} \sin \left(2 x\right) {|}_{0}^{1}$
$\implies I = - {10}^{4} {e}^{- 2} \sin 2$