# How do you find the definite integral of 2x* (x-3)^5 dx from [-1, 2]?

Jul 2, 2016

$= \frac{4101}{7}$

#### Explanation:

we'll use IBP

$\int u v ' = u v - \int u ' v$

so for
${\int}_{- 1}^{2} \mathrm{dx} q \quad 2 x {\left(x - 3\right)}^{5}$

we can set

$u = 2 x , u ' = 2$
$v ' = {\left(x - 3\right)}^{5} , v = \frac{1}{6} {\left(x - 3\right)}^{6}$

$\implies \left[2 x \cdot \frac{1}{6} {\left(x - 3\right)}^{6}\right] {\setminus}_{- 1}^{2} - {\int}_{- 1}^{2} \mathrm{dx} q \quad 2 \cdot \frac{1}{6} {\left(x - 3\right)}^{6}$

$= \left[\frac{1}{3} x {\left(x - 3\right)}^{6}\right] {\setminus}_{- 1}^{2} - \frac{1}{3} {\int}_{- 1}^{2} \mathrm{dx} q \quad {\left(x - 3\right)}^{6}$

$= \left[\frac{1}{3} x {\left(x - 3\right)}^{6} - \frac{1}{3} \cdot \frac{1}{7} {\left(x - 3\right)}^{7}\right] {\setminus}_{- 1}^{2}$

$= \left[\frac{1}{3} x {\left(x - 3\right)}^{6} - \frac{1}{21} {\left(x - 3\right)}^{7}\right] {\setminus}_{- 1}^{2}$

$= \frac{2}{3} + \frac{1}{21} + {4}^{6} / 3 - {4}^{7} / 21$

$= \frac{4101}{7}$