# What is the antiderivative of (16 - x)^(3/2)?

Jan 29, 2016

$- \frac{2}{5} {\left(16 - x\right)}^{\frac{3}{2}} + C$

#### Explanation:

Antiderivative is just the integration. For handling this type of problem, you should be comfortable with the power rule.

color(red)(intquad x^n dx = x^(n+1)/(n+1)+C

Our problem we have to find

$\int \quad {\left(16 - x\right)}^{\frac{3}{2}} \mathrm{dx}$

We shall use a $u$ substitution just to make into a form which we are comfortable with.

Let $16 - x = u$
Differentiating with respect to $x$ on both the sides we get.

$- \mathrm{dx} = \mathrm{du}$
$\mathrm{dx} = - \mathrm{du}$

Our integral now becomes

$\int \quad {u}^{\frac{3}{2}} \left(- \mathrm{du}\right)$
$= - \int \quad {u}^{\frac{3}{2}} \mathrm{du}$

Use the power rule

$= - {u}^{\frac{3}{2} + 1} / \left(\frac{3}{2} + 1\right) + C$
$= - {u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) + C$
$= - \frac{2}{5} {u}^{\frac{5}{2}} + C$

Substituting back for $u$ we get,

$= - \frac{2}{5} {\left(16 - x\right)}^{\frac{3}{2}} + C$