What is the integral of (xsqrt(9-x^2)) ?

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

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integral of (xsqrt(9-x^2))

\int (x 9 - x^{2}) d x

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

Solution steps

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

Apply u-substitution

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

= \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{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 9 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = e^{2 y + 2 x}

\frac{d}{d x} (x - 3 x^{2})

4 x^{2} y^{2} d x + x^{3} y d y = 0

\int_{- 3}^{3} (9 - x^{2})^{2} d x

\int e^{- 2 x} (x^{2} + 5) d x

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