# The force applied against a moving object travelling on a linear path is given by F(x)= x^2+xe^x . How much work would it take to move the object over x in [4, 5] ?

Dec 24, 2016

The answer is $= 450.16 J$

#### Explanation:

The work done by a force $F \left(x\right)$ from ${x}_{1}$ to ${x}_{2}$ is

$W = {\int}_{{x}_{1}}^{{x}_{2}} F \left(x\right) \mathrm{dx}$

Here $F \left(x\right) = {x}^{2} + x {e}^{x}$

${x}_{1} = 4$ and ${x}_{2} = 5$

So,

$W = {\int}_{4}^{5} \left({x}^{2} + x {e}^{x}\right) \mathrm{dx}$

$= {\int}_{4}^{5} {x}^{2} \mathrm{dx} + {\int}_{4}^{5} x {e}^{x} \mathrm{dx}$

$= {\left[{x}^{3} / 3\right]}_{4}^{5} + {\int}_{4}^{5} x {e}^{x} \mathrm{dx}$

We perform the second inegration by parts

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

$u = x$, $\implies$, $u ' = 1$

$v ' = {e}^{x}$, $\implies$, $v = {e}^{x}$

$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \mathrm{dx}$

$= x {e}^{x} - {e}^{x}$

So,
$W = {\left[{x}^{3} / 3\right]}_{4}^{5} + {\left[x {e}^{x} - {e}^{x}\right]}_{4}^{5}$

$= \left({5}^{3} / 3 - {4}^{3} / 3\right) + \left(5 {e}^{5} - {e}^{5} - 4 {e}^{4} + {e}^{4}\right)$

$= \frac{61}{3} + {e}^{4} \left(4e-3\right)$

$= 450.16 J$