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

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

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

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{0} 1 + cos (2 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

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

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

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

= \frac{1}{2} (\frac{π}{2} + 0)

Simplify

\frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{4}

= \frac{π}{4}

\int_{\frac{π}{4}}^{\frac{π}{2}} sec^{2} (x) d x

\int_{0}^{1} \frac{1}{2 x + 1} d x

\int_{- 1}^{1} sin (2 x) + cos (3 x) d x

\int_{1}^{2} 4 e^{3 x - 2} d x

\int_{3}^{10} 1

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