What is the integral of xsqrt(36-x^2) ?

The integral of xsqrt(36-x^2) is -1/3 (36-x^2)^{3/2}+C

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integral of xsqrt(36-x^2)

\int x 36 - x^{2} d x

- \frac{1}{3} (36 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 36 - 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 = 36 - x^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (36 - x^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (36 - x^{2})^{\frac{3}{2}} : - \frac{1}{3} (36 - x^{2})^{\frac{3}{2}}

= - \frac{1}{3} (36 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (36 - x^{2})^{\frac{3}{2}} + C

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\int 3 x - 2 d x

\frac{d y}{d x} = 3 y \cdot e^{x + 3}

y = \frac{x ^{2} + 7}{5 x - x ^{2}}, 1, 4

\frac{\partial}{\partial x} (r sin (x))

What is the integral of xsqrt(36-x^2) ?
The integral of xsqrt(36-x^2) is -1/3 (36-x^2)^{3/2}+C