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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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 \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} (\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} (π + 0)

Simplify

\frac{1}{2} (π + 0) : \frac{π}{2}

= \frac{π}{2}

\int_{4}^{8} (- 6 x + 5) d x

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

\int_{0}^{2 π} sin (x) \cdot cos (x) d x

\int_{- \frac{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} d x

\int_{0}^{y} \frac{1}{2 y} d y

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