What is the integral from 2 to 4 of sqrt(16-r^2)r ?

The integral from 2 to 4 of sqrt(16-r^2)r is 8sqrt(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 4 of sqrt(16-r^2)r

\int_{2}^{4} 16 - r^{2} r d r

83

Decimal Notation

13.85640 \dots

Solution steps

\int_{2}^{4} 16 - r^{2} r d r

Apply u-substitution

: \int_{12}^{0} - \frac{u}{2} d u

= \int_{12}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{12} - \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}^{12} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

Compute the boundaries:

163

= \frac{1}{2} \cdot 163

Simplify

= 83

\int_{1}^{6} (\frac{9}{x} - e^{- x}) d x

\int_{0}^{3} (x^{3} - 4 x) d x

\int_{0}^{3} \frac{3}{x + 4} d x

\int_{1}^{e} \frac{1}{x} ln (x) d x

\int_{0}^{5} e^{- x^{2}} d x

What is the integral from 2 to 4 of sqrt(16-r^2)r ?
The integral from 2 to 4 of sqrt(16-r^2)r is 8sqrt(3)