# The force applied against a moving object travelling on a linear path is given by F(x)= cosx + 2 . How much work would it take to move the object over x in [ 0, (13 pi) / 6] ?

Jul 22, 2017

The work is $= 14.1 J$

#### Explanation:

$\int \cos x \mathrm{dx} = \sin x + C$

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

The work is

$\Delta W = F \left(x\right) \cdot \Delta x$

$F \left(x\right) = 2 + \cos x$

$W = {\int}_{0}^{\frac{13}{6} \pi} \left(2 + \cos x\right) \mathrm{dx}$

$= {\left[2 x + \sin x\right]}_{0}^{\frac{13}{6} \pi}$

$= \left(2 \cdot \frac{13}{6} \pi + \sin \left(\frac{13}{6} \pi\right)\right) - \left(0 + \sin 0\right)$

$= \frac{13}{3} \pi + \sin \left(\frac{13}{6} \pi\right)$

$= \frac{31}{3} \pi + \frac{1}{2}$

$= 14.1 J$