What is the integral from 0 to 2 of sqrt(4-r^2)r ?

The integral from 0 to 2 of sqrt(4-r^2)r is 8/3

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(4-r^2)r

\int_{0}^{2} 4 - r^{2} r d r

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{0}^{2} 4 - r^{2} r d r

Apply u-substitution

: \int_{4}^{0} - \frac{u}{2} d u

= \int_{4}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{4} - \frac{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}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{1}{2} \cdot \frac{16}{3}

Simplify

= \frac{8}{3}

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

\int_{- 5}^{1} - x^{2} - 3 x d x

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

\int_{1}^{x} \frac{1}{1 + t ^{3}} d t

\int_{- x}^{x} x + y d y

What is the integral from 0 to 2 of sqrt(4-r^2)r ?
The integral from 0 to 2 of sqrt(4-r^2)r is 8/3