What is the integral from 0 to 1 of xsqrt(2x-x^2) ?

The integral from 0 to 1 of xsqrt(2x-x^2) is -1/3+pi/4

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integral from 0 to 1 of xsqrt(2x-x^2)

\int_{0}^{1} x 2 x - x^{2} d x

- \frac{1}{3} + \frac{π}{4}

Decimal Notation

0.45206 \dots

Solution steps

\int_{0}^{1} x 2 x - x^{2} d x

Complete the square

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

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

Apply u-substitution

: \int_{- 1}^{0} (u + 1) - u^{2} + 1 d u

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

Expand

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

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

\int_{- 1}^{0} u - u^{2} + 1 d u = - \frac{1}{3}

\int_{- 1}^{0} - u^{2} + 1 d u = \frac{π}{4}

= - \frac{1}{3} + \frac{π}{4}

\int_{- 5}^{5} (- 0.4 x^{2} + 10) d x

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

What is the integral from 0 to 1 of xsqrt(2x-x^2) ?
The integral from 0 to 1 of xsqrt(2x-x^2) is -1/3+pi/4