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

The integral from 0 to 4 of sqrt(4r^2+1)r is (65sqrt(65)-1)/(12)

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

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

\frac{65 65 - 1}{12}

Decimal

43.58722 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{65} \frac{u}{8} d u

= \int_{1}^{65} \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}^{65} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{65}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{65}

Compute the boundaries:

\frac{130 65}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{130 65}{3} - \frac{2}{3})

Simplify

= \frac{65 65 - 1}{12}

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What is the integral from 0 to 4 of sqrt(4r^2+1)r ?
The integral from 0 to 4 of sqrt(4r^2+1)r is (65sqrt(65)-1)/(12)