# 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 [2, 5] ?

Jan 22, 2016

$4 {e}^{5} - {e}^{2} + 39$

#### Explanation:

For constant force

$W = F \cdot x$

For non-constant force

$\mathrm{dW} = F \cdot \mathrm{dx}$

Integrate by-parts

$W = {\int}_{2}^{5} F \cdot \mathrm{dx}$

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

$= {\int}_{2}^{5} {x}^{2} \cdot \mathrm{dx} + {\int}_{2}^{5} x \frac{d}{\mathrm{dx}} \left({e}^{x}\right) \cdot \mathrm{dx}$

$= {\left[{x}^{3} / 3\right]}_{2}^{5} + {\left[x {e}^{x}\right]}_{2}^{5} - {\int}_{2}^{5} {e}^{x} \frac{d}{\mathrm{dx}} \left(x\right) \cdot \mathrm{dx}$

$= \left[{5}^{3} / 3 - {2}^{3} / 3\right] + \left[5 {e}^{5} - 2 {e}^{2}\right] - {\int}_{2}^{5} {e}^{x} \cdot \mathrm{dx}$

$= 39 + \left(5 {e}^{3} - 2\right) {e}^{2} - {\left[{e}^{x}\right]}_{2}^{5}$

$= 39 + \left(5 {e}^{3} - 2\right) {e}^{2} - \left({e}^{3} - 1\right) {e}^{2}$

$= 39 + \left(4 {e}^{3} - 1\right) {e}^{2}$