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

The integral from 0 to 2 of x^2sqrt(2x-x^2) is (5pi)/8

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

\int_{0}^{2} x^{2} 2 x - x^{2} d x

\frac{5 π}{8}

Decimal

1.96349 \dots

Solution steps

\int_{0}^{2} x^{2} 2 x - x^{2} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

= \int_{- 1}^{1} u^{2} - u^{2} + 1 + 2 u - u^{2} + 1 + - u^{2} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} u^{2} - u^{2} + 1 d u + \int_{- 1}^{1} 2 u - u^{2} + 1 d u + \int_{- 1}^{1} - u^{2} + 1 d u

\int_{- 1}^{1} u^{2} - u^{2} + 1 d u = \frac{π}{8}

\int_{- 1}^{1} 2 u - u^{2} + 1 d u = 0

\int_{- 1}^{1} - u^{2} + 1 d u = \frac{π}{2}

= \frac{π}{8} + 0 + \frac{π}{2}

Simplify

\frac{π}{8} + 0 + \frac{π}{2} : \frac{5 π}{8}

= \frac{5 π}{8}

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What is the integral from 0 to 2 of x^2sqrt(2x-x^2) ?
The integral from 0 to 2 of x^2sqrt(2x-x^2) is (5pi)/8