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

The integral from 0 to 2 of-2x^2sqrt(2x-x^2) is -(5pi)/4

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

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

- \frac{5 π}{4}

Decimal Notation

- 3.92699 \dots

Solution steps

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

Take the constant out:

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

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

Complete the square

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

= - 2 \cdot \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

= - 2 \cdot \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

= - 2 \cdot \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

= - 2 (\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}

= - 2 (\frac{π}{8} + 0 + \frac{π}{2})

Simplify

- 2 (\frac{π}{8} + 0 + \frac{π}{2}) : - \frac{5 π}{4}

= - \frac{5 π}{4}

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What is the integral from 0 to 2 of-2x^2sqrt(2x-x^2) ?
The integral from 0 to 2 of-2x^2sqrt(2x-x^2) is -(5pi)/4