What is the integral from 0 to 2 of ysqrt(2y-y^2) ?

The integral from 0 to 2 of ysqrt(2y-y^2) is pi/2

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integral from 0 to 2 of ysqrt(2y-y^2)

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Complete the square

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

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

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}

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