What is the integral of ysqrt(9-y^2) ?

The integral of ysqrt(9-y^2) is -1/3 (9-y^2)^{3/2}+C

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integral of ysqrt(9-y^2)

\int y 9 - y^{2} d y

- \frac{1}{3} (9 - y^{2})^{\frac{3}{2}} + C

Solution steps

\int y 9 - y^{2} d y

Apply u-substitution

= \int - \frac{u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int u d u

Apply the Power Rule

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 - y^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (9 - y^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (9 - y^{2})^{\frac{3}{2}} : - \frac{1}{3} (9 - y^{2})^{\frac{3}{2}}

= - \frac{1}{3} (9 - y^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (9 - y^{2})^{\frac{3}{2}} + C

What is the integral of ysqrt(9-y^2) ?
The integral of ysqrt(9-y^2) is -1/3 (9-y^2)^{3/2}+C