What is the integral from 0 to 2 of sqrt((2x-x^2)^3) ?

The integral from 0 to 2 of sqrt((2x-x^2)^3) is (3pi)/8

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integral from 0 to 2 of sqrt((2x-x^2)^3)

\int_{0}^{2} (2 x - x^{2})^{3} d x

\frac{3 π}{8}

Decimal

1.17809 \dots

Solution steps

\int_{0}^{2} (2 x - x^{2})^{3} d x

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

= \int_{0}^{2} (- (x - 1)^{2} + 1)^{3} d x

Apply u-substitution

: \int_{- 1}^{1} (- u^{2} + 1)^{\frac{3}{2}} d u

= \int_{- 1}^{1} (- u^{2} + 1)^{\frac{3}{2}} d u

Apply Integration By Parts

: [u (- u^{2} + 1)^{\frac{3}{2}} - \int - 3 u^{2} - u^{2} + 1 d u]_{- 1}^{1}

= [u (- u^{2} + 1)^{\frac{3}{2}} - \int - 3 u^{2} - u^{2} + 1 d u]_{- 1}^{1}

\int - 3 u^{2} - u^{2} + 1 d u = - \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

= [u (- u^{2} + 1)^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u))))]_{- 1}^{1}

Simplify

= [u (- u^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))]_{- 1}^{1}

Compute the boundaries:

\frac{3 π}{8}

= \frac{3 π}{8}

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What is the integral from 0 to 2 of sqrt((2x-x^2)^3) ?
The integral from 0 to 2 of sqrt((2x-x^2)^3) is (3pi)/8