# How do you find the definite integral of (x) / sqrt(4 + 3x) dx  from [0, 7]?

##### 1 Answer
Oct 30, 2016

Please see the explanation section below.

#### Explanation:

${\int}_{0}^{7} \frac{x}{\sqrt{4 + 3 x}} \mathrm{dx}$

Let $u = 4 + 3 x$.

This makes $\mathrm{du} = 3 \mathrm{dx}$, and $x = \frac{u - 4}{3}$ and the limits of integration become $u = 4$ to $u = 25$

Substitute

$\frac{1}{3} {\int}_{0}^{7} x {\left(4 + 3 x\right)}^{- \frac{1}{2}} \left(3 \mathrm{dx}\right) = \frac{1}{3} {\int}_{4}^{25} \frac{u - 4}{3} \cdot {u}^{- \frac{1}{2}} \mathrm{du}$

$= \frac{1}{9} {\int}_{4}^{25} \left({u}^{\frac{1}{2}} - 4 {u}^{- \frac{1}{2}}\right) \mathrm{du}$

$= \frac{1}{9} {\left[\frac{2}{3} {u}^{\frac{3}{2}} - 8 {u}^{\frac{1}{2}}\right]}_{4}^{25}$

$= \frac{1}{9} {\left[\frac{2}{3} {u}^{\frac{3}{2}} - \frac{24}{3} {u}^{\frac{1}{2}}\right]}_{4}^{25}$

$= \frac{2}{27} {\left[\sqrt{u} \left(u - 12\right)\right]}_{4}^{25}$

$= \frac{2}{27} \left[5 \left(13\right) - 2 \left(- 8\right)\right]$

$= \frac{2}{27} \left(65 + 16\right) = \frac{2}{27} \left(81\right) = 6$