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

The integral from 0 to 1 of 2xsqrt(1-x) is 8/15

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

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

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

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

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

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

= 2 (- (- (- \frac{2}{5} + \frac{2}{3})))

Simplify

2 (- (- (- \frac{2}{5} + \frac{2}{3}))) : \frac{8}{15}

= \frac{8}{15}

What is the integral from 0 to 1 of 2xsqrt(1-x) ?
The integral from 0 to 1 of 2xsqrt(1-x) is 8/15