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

The integral from 0 to 1 of sqrt(x)sqrt(1-x) is pi/8

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integral from 0 to 1 of sqrt(x)sqrt(1-x)

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

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} sin^{2} (v) cos^{2} (v) d v

= 2 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (v) cos^{2} (v) d v

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 v )}{8} d v

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d v - \int_{0}^{\frac{π}{2}} cos (4 v) d v)

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 v) d v = 0

= 2 \cdot \frac{1}{8} (\frac{π}{2} - 0)

Simplify

2 \cdot \frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{8}

= \frac{π}{8}

\int_{0}^{2} r e^{- r^{2}} d r

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What is the integral from 0 to 1 of sqrt(x)sqrt(1-x) ?
The integral from 0 to 1 of sqrt(x)sqrt(1-x) is pi/8