# What is the volume of the solid produced by revolving f(x)=xe^x-(x/2)e^x, x in [2,7] around the x-axis?

Nov 28, 2016

The volume of the solid produced by revolving around the x-axis the trapezoid of the function $f \left(x\right)$ is calculated using the formula:

$V = \pi {\int}_{a}^{b} {f}^{2} \left(x\right) \mathrm{dx}$

#### Explanation:

$V = \pi {\int}_{2}^{7} {\left(x {e}^{x} - \left(\frac{x}{2}\right) {e}^{x}\right)}^{2} \mathrm{dx}$

$V = \pi {\int}_{2}^{7} \left({x}^{2} {e}^{2 x} - 2 x {e}^{x} \left(\frac{x}{2}\right) {e}^{x} + {x}^{2} / 4 {e}^{2 x}\right) \mathrm{dx}$

$V = \pi {\int}_{2}^{7} \left({x}^{2} {e}^{2 x} - {x}^{2} {e}^{2 x} + {x}^{2} / 4 {e}^{2 x}\right) \mathrm{dx} = \pi {\int}_{2}^{7} {x}^{2} / 4 {e}^{2 x} \mathrm{dx}$

Substitute $2 x = t$:

$V = \frac{\pi}{32} {\int}_{1}^{\frac{7}{2}} {t}^{2} {e}^{t} \mathrm{dt}$

The integral is solved iteratively by parts:

$\int {t}^{2} {e}^{t} \mathrm{dt} = \int {t}^{2} d \left({e}^{t}\right) = {t}^{2} {e}^{t} - \int 2 t {e}^{t} \mathrm{dt} = {t}^{2} {e}^{t} - 2 t {e}^{t} + 2 \int {e}^{t} \mathrm{dt} = {e}^{t} \left({t}^{2} - 2 t + 2\right)$

Finally:

$V = \frac{\pi}{32} \left[{e}^{\frac{7}{2}} \left(\frac{49}{4} - 7 + 2\right) - e \left(1 - 2 + 2\right)\right] = \frac{\pi}{32} \left[{e}^{\frac{7}{2}} \frac{49 - 28 + 8}{4} - e\right] = \frac{\pi \cdot e}{32} \left(\frac{29}{4} {e}^{\frac{5}{2}} - 1\right)$