How would you integrate something like

sqrt(9-y)*y dy from 0 to 9

Printable View

- Sep 6th 2012, 01:42 PMCoffeeBirdHow to integrate this problem
How would you integrate something like

sqrt(9-y)*y dy from 0 to 9 - Sep 6th 2012, 01:51 PMHallsofIvyRe: How to integrate this problem
The problem is that "9- y" inside the square root, right? So try u= 9- y. Then du= -dy. and, of course, y= 9- u. When y= 0, u= 9 and when y= 9, u= 0.

So the $\displaystyle \sqrt{9- y}y dy= \sqrt{u}(9- u)(-du)= -(9u^{1/2}- u^{3/2))du$ and the integral will be from 9 to 0. Swapping the limits of integration gets rid of that negative. - Sep 6th 2012, 02:09 PMCoffeeBirdRe: How to integrate this problem
wow, that was very clearly explained, thank you

- Sep 6th 2012, 03:46 PMSorobanRe: How to integrate this problem
Hello, CoffeeBird!

Quote:

$\displaystyle \int^9_0y\sqrt{9-y}\:dy$

If the expression under the radical is,*linear*

. . we can let $\displaystyle u$ equal the entire radical.

Let $\displaystyle u \:=\:\sqrt{9-y} \quad\Rightarrow\quad u^2 \:=\:9-y \quad\Rightarrow\quad y\:=\:9-u^2 \quad\Rightarrow\quad du \:=\:-2u\,du$

Substitute: .$\displaystyle \int (9-u^2)\cdot u \cdot (-2u\,du) \;=\; -2\int(9u^2 - u^4)\,du$

. . . . . . . . $\displaystyle =\;-2\left(3u^3 - \tfrac{1}{5}u^5\right) + C \;=\;-\tfrac{2}{5}u^3\left(15 - u^2\right) + C$

Back-substitute: .$\displaystyle -\tfrac{2}{5}(9-y)^{\frac{3}{2}}(15 - [9-y]) + C \;=\;-\tfrac{2}{5}(9-y)^{\frac{3}{2}}(6 + y) + C$

Evaluate: .$\displaystyle -\tfrac{2}{5}(9-y)^{\frac{3}{2}}(6+y)\,\bigg]^9_0 \;=\;-\tfrac{2}{5}(0)^{\frac{3}{2}}(15) \:+\:\tfrac{2}{5}(9)^{\frac{3}{2}}(6) $

. . . . . . . . $\displaystyle =\;0 + \tfrac{2}{5}(27)(6) \;=\;\frac{324}{5}$