# How do you integrate by substitution int (9-y)sqrty dy?

##### 2 Answers
Mar 19, 2018

The answer is $= 6 {y}^{\frac{3}{2}} - \frac{2}{5} {y}^{\frac{5}{2}} + C$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(x \ne - 1\right)$

Let $u = \sqrt{y}$, $\implies$, $\mathrm{du} = \frac{\mathrm{dy}}{2 \sqrt{y}}$

Therefore,

$\int \left(9 - y\right) \sqrt{y} \mathrm{dy} = \int \left(9 - {u}^{2}\right) \sqrt{y} \cdot 2 \sqrt{y} \mathrm{du}$

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

$= 2 \cdot 9 {u}^{3} / 3 - \frac{2}{5} {u}^{5}$

$= 6 {y}^{\frac{3}{2}} - \frac{2}{5} {y}^{\frac{5}{2}} + C$

Mar 19, 2018

$\frac{2}{5} {y}^{\frac{3}{2}} \left(15 - y\right) + c$

#### Explanation:

$I = \int \left(9 - y\right) \sqrt{y} \mathrm{dy}$

to do this by substitution:

u=sqrty=>u=y^(1/2

$\implies \mathrm{du} = \frac{1}{2} {y}^{- \frac{1}{2}} \mathrm{dy}$

$\mathrm{dy} = 2 {y}^{\frac{1}{2}} \mathrm{du}$

$I = \int \left(9 - {u}^{2}\right) {y}^{\frac{1}{2}} 2 {y}^{\frac{1}{2}} \mathrm{du}$

$= \int \left(9 - {u}^{2}\right) y \mathrm{du}$

$= 2 \int \left(9 - {u}^{2}\right) {u}^{2} \mathrm{du}$

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

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

$= \frac{2}{5} {u}^{3} \left(15 - {u}^{2}\right) = c$

$= \frac{2}{5} {y}^{\frac{3}{2}} \left(15 - y\right) + c$

note for this integral it would be more efficient to multiply the brackets out and integrate directly using the power rule

ie

$\int \left(9 - y\right) {y}^{\frac{1}{2}} \mathrm{dy} = \int \left(9 {y}^{\frac{1}{2}} - {y}^{\frac{3}{2}}\right) \mathrm{dy}$

etc