# How do you integrate int y^2sqrtydy?

Jan 4, 2017

$\int \setminus {y}^{2} \sqrt{y} \setminus \mathrm{dy} = \frac{2}{7} {y}^{3} \sqrt{y} + c$

#### Explanation:

Using fractional powers, and the rule of indices ${a}^{m} {a}^{n} = {a}^{m + n}$ we can write:

$\int \setminus {y}^{2} \sqrt{y} \setminus \mathrm{dy} = \int \setminus {y}^{2} \setminus {y}^{\frac{1}{2}} \setminus \mathrm{dy}$
$\text{ } = \int \setminus {y}^{\frac{5}{2}} \setminus \mathrm{dy}$

Then using the power rule for integration $\int {x}^{n} \setminus \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ we have:

$\int \setminus {y}^{2} \sqrt{y} \setminus \mathrm{dy} = {y}^{\frac{7}{2}} / \left(\frac{7}{2}\right) + c$
$\text{ } = \frac{2}{7} {y}^{\frac{7}{2}} + c$
$\text{ } = \frac{2}{7} {y}^{3} \sqrt{y} + c$