What is the integral from 0 to 2 of sqrt(1+4r^2)r ?

The integral from 0 to 2 of sqrt(1+4r^2)r is (17sqrt(17)-1)/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2 of sqrt(1+4r^2)r

\int_{0}^{2} 1 + 4 r^{2} r d r

\frac{17 17 - 1}{12}

Decimal

5.75773 \dots

Solution steps

\int_{0}^{2} 1 + 4 r^{2} r d r

Apply u-substitution

= \int_{1}^{17} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{17} u d u

Apply the Power Rule

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{17 17 - 1}{12}

What is the integral from 0 to 2 of sqrt(1+4r^2)r ?
The integral from 0 to 2 of sqrt(1+4r^2)r is (17sqrt(17)-1)/(12)