# Evaluate the Integral ?

## ${\int}_{1}^{2} \left(\frac{{v}^{3} - 3 {v}^{6}}{v} ^ 4\right) \mathrm{dv}$ I got to the point where I used the properties of integrals and my answer was $\ln 2 + 5$. Where F(b)-F(a). The answer was $\ln 2 + 7$. Any ideas where I went wrong. I don't see where they are getting 7 from.

Aug 11, 2018

$\ln 2 - 7$.

#### Explanation:

${\int}_{1}^{2} \left(\frac{{v}^{3} - 3 {v}^{6}}{v} ^ 4\right) \mathrm{dv}$,

$= {\int}_{1}^{2} \left({v}^{3} / {v}^{4} - 3 {v}^{6} / {v}^{4}\right) \mathrm{dv}$,

$= {\int}_{1}^{2} \left(\frac{1}{v} - 3 {v}^{2}\right) \mathrm{dv}$,

$= {\left[\ln | v | - 3 \cdot {v}^{2 + 1} / \left(2 + 1\right)\right]}_{1}^{2}$,

$= {\left[\ln | v | - {v}^{3}\right]}_{1}^{2}$,

$= \left[\left\{\ln | 2 | - {2}^{3}\right\} - \left\{\ln | 1 | - {1}^{3}\right\}\right]$,

$= \ln 2 - 8 - 0 + 1$.

$\Rightarrow {\int}_{1}^{2} \left(\frac{{v}^{3} - 3 {v}^{6}}{v} ^ 4\right) \mathrm{dv} = \ln 2 - 7$.