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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \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}

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