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

The integral from 0 to 1 of sqrt(x-x^2) is pi/8

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

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

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

Complete the square

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{2} cos^{2} (v) d v

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

Take the constant out:

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

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

Rewrite using trig identities

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

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

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

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

Simplify

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

= \frac{π}{8}

\int (y - 1) (2 y + 1) d y

\frac{d y}{d x} + 2 x y = 2 x^{3}

\frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}

x \to \infty lim (3^{x} - x)

n = 1 \sum \infty \frac{n}{3 n + 2}

What is the integral from 0 to 1 of sqrt(x-x^2) ?
The integral from 0 to 1 of sqrt(x-x^2) is pi/8