What is the integral from 0 to 1 of xsqrt(4x-x^2) ?

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

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integral from 0 to 1 of xsqrt(4x-x^2)

\int_{0}^{1} x 4 x - x^{2} d x

4 \cdot \frac{π}{3} - 23

Decimal Notation

0.72468 \dots

Solution steps

\int_{0}^{1} x 4 x - x^{2} d x

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

= \int_{0}^{1} x - (x - 2)^{2} + 4 d x

Apply u-substitution

: \int_{- 2}^{- 1} (u + 2) - u^{2} + 4 d u

= \int_{- 2}^{- 1} (u + 2) - u^{2} + 4 d u

Expand

(u + 2) - u^{2} + 4 : u - u^{2} + 4 + 2 - u^{2} + 4

= \int_{- 2}^{- 1} u - u^{2} + 4 + 2 - u^{2} + 4 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_{- 2}^{- 1} u - u^{2} + 4 d u + \int_{- 2}^{- 1} 2 - u^{2} + 4 d u

\int_{- 2}^{- 1} u - u^{2} + 4 d u = - 3

\int_{- 2}^{- 1} 2 - u^{2} + 4 d u = 4 \cdot \frac{π}{3} - 3

= - 3 + 4 \cdot \frac{π}{3} - 3

Simplify

= 4 \cdot \frac{π}{3} - 23

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