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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

= 0 + \frac{π}{2}

Simplify

= \frac{π}{2}

x \to 0 + lim (ln^{2} (x) - ln (x))

\frac{d}{d x} (\frac{sec ^{2} ( x ^{3} )}{2})

\frac{\partial}{\partial x} (x sin (5 x^{2} y))

\int - 2 sin^{5} (x) cos^{2} (x) d x

\frac{d}{d x} (1.795 x^{2})

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