What is the integral of 8xsqrt(40-2x^2) ?

The integral of 8xsqrt(40-2x^2) is -4/3 (-2x^2+40)^{3/2}+C

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integral of 8xsqrt(40-2x^2)

\int 8 x 40 - 2 x^{2} d x

- \frac{4}{3} (- 2 x^{2} + 40)^{\frac{3}{2}} + C

Solution steps

\int 8 x 40 - 2 x^{2} d x

Take the constant out:

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

= 8 \cdot \int x 40 - 2 x^{2} d x

Apply u-substitution

: \int - \frac{u}{4} d u

= 8 \cdot \int - \frac{u}{4} d u

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int u d u)

Apply the Power Rule

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

= 8 (- \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 40 - 2 x^{2}

= 8 (- \frac{1}{4} \cdot \frac{2}{3} (40 - 2 x^{2})^{\frac{3}{2}})

Simplify

8 (- \frac{1}{4} \cdot \frac{2}{3} (40 - 2 x^{2})^{\frac{3}{2}}) : - \frac{4}{3} (- 2 x^{2} + 40)^{\frac{3}{2}}

= - \frac{4}{3} (- 2 x^{2} + 40)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{4}{3} (- 2 x^{2} + 40)^{\frac{3}{2}} + C

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What is the integral of 8xsqrt(40-2x^2) ?
The integral of 8xsqrt(40-2x^2) is -4/3 (-2x^2+40)^{3/2}+C