What is the integral from 0 to 1/2 of rsqrt(1-r^2) ?

The integral from 0 to 1/2 of rsqrt(1-r^2) is 1/3-(sqrt(3))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/2 of rsqrt(1-r^2)

\int_{0}^{\frac{1}{2}} r 1 - r^{2} d r

\frac{1}{3} - \frac{3}{8}

Decimal

0.11682 \dots

Solution steps

\int_{0}^{\frac{1}{2}} r 1 - r^{2} d r

Apply u-substitution

: \int_{1}^{\frac{3}{2}} - u^{2} d u

= \int_{1}^{\frac{3}{2}} - u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{3}{2}}^{1} - u^{2} d u

Take the constant out:

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

= - (- \int_{\frac{3}{2}}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1}

= - (- [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1})

Simplify

= [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1}

Compute the boundaries:

\frac{1}{3} - \frac{3}{8}

= \frac{1}{3} - \frac{3}{8}

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

\int_{- \infty}^{x} e^{- x} d x

\int_{2}^{\infty} u e^{- u} d u

\int_{0}^{3} \frac{1}{3} x^{2} d x

\int_{0}^{2} (x^{4}) d x

What is the integral from 0 to 1/2 of rsqrt(1-r^2) ?
The integral from 0 to 1/2 of rsqrt(1-r^2) is 1/3-(sqrt(3))/8