What is the integral of ysqrt(9-y) ?

The integral of ysqrt(9-y) is 2/5 (9-y)^{5/2}-6(9-y)^{3/2}+C

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integral of ysqrt(9-y)

\int y 9 - y d y

\frac{2}{5} (9 - y)^{\frac{5}{2}} - 6 (9 - y)^{\frac{3}{2}} + C

Solution steps

\int y 9 - y d y

Apply u-substitution

= \int - (- u + 9) u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int (- u + 9) u d u

Expand

(- u + 9) u : - u^{\frac{3}{2}} + 9 u

= - \int - u^{\frac{3}{2}} + 9 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{\frac{3}{2}} d u + \int 9 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 9 - y

= - (- \frac{2}{5} (9 - y)^{\frac{5}{2}} + 6 (9 - y)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (9 - y)^{\frac{5}{2}} - 6 (9 - y)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (9 - y)^{\frac{5}{2}} - 6 (9 - y)^{\frac{3}{2}} + C

What is the integral of ysqrt(9-y) ?
The integral of ysqrt(9-y) is 2/5 (9-y)^{5/2}-6(9-y)^{3/2}+C