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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (1-x^2)^{1/2}

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= \frac{1}{2} (\frac{π}{2} + 0)

Simplify

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

= \frac{π}{4}

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

\int_{3 x^{2} - 6 x}^{2 x - x^{2}} 3 x d x

\int_{0}^{1} x sin (\frac{nπ}{2} x) d x

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

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 3)^{2} d x

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