What is the integral of xsqrt(6-x^2) ?

The integral of xsqrt(6-x^2) is -1/3 (6-x^2)^{3/2}+C

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integral of xsqrt(6-x^2)

\int x 6 - x^{2} d x

- \frac{1}{3} (6 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 6 - 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 = 6 - x^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (6 - x^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (6 - x^{2})^{\frac{3}{2}} : - \frac{1}{3} (6 - x^{2})^{\frac{3}{2}}

= - \frac{1}{3} (6 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (6 - x^{2})^{\frac{3}{2}} + C

f (x) = 4 x^{2} + 3 x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (9 x)

\frac{1}{X ^{2}}

\frac{4 ( - s + 16 )}{s ^{2} + 10 s}

(- 4.5) (2.3)

What is the integral of xsqrt(6-x^2) ?
The integral of xsqrt(6-x^2) is -1/3 (6-x^2)^{3/2}+C