# The force applied against a moving object travelling on a linear path is given by F(x)=3x^2+e^x . How much work would it take to move the object over x in [0, 3 ] ?

Sep 10, 2017

The work is $= 46.1 J$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

$\int {e}^{x} \mathrm{dx} = {e}^{x} + C$

The work is

$\Delta W = F \Delta x$

$\Delta W = \left(3 {x}^{2} + {e}^{x}\right) \Delta x$

$W = {\int}_{0}^{3} \left(3 {x}^{2} + {e}^{x}\right) \mathrm{dx}$

$= {\left[{x}^{3} + {e}^{x}\right]}_{0}^{3}$

$= \left({3}^{3} + {e}^{3}\right) - \left(0 + {e}^{0}\right)$

$= 27 + {e}^{3} - 1$

$= 46.1 J$

Sep 10, 2017

I got $46 J$.

#### Explanation:

Here you have a variable force so for an infinitesimal (=very small) Work you have:

$\mathrm{dW} = F \left(x\right) \mathrm{dx}$

integrating:

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

$W = {\int}_{0}^{3} \left(3 {x}^{2} + {e}^{x}\right) \mathrm{dx} = \cancel{3} {x}^{3} / \cancel{3} + {e}^{x} = {x}^{3} + {e}^{x} {|}_{0}^{3} =$
$= \left({3}^{3} + {e}^{3}\right) - \left({0}^{3} + {e}^{0}\right) = 46 J$