What is the integral from 0 to 3 of sqrt(1+4r^2)r ?

The integral from 0 to 3 of sqrt(1+4r^2)r is (37sqrt(37)-1)/(12)

integral from 0 to 3 of sqrt(1+4r^2)r

\int_{0}^{3} 1 + 4 r^{2} r d r

\frac{37 37 - 1}{12}

Decimal

18.67185 \dots

Solution steps

\int_{0}^{3} 1 + 4 r^{2} r d r

Apply u-substitution

= \int_{1}^{37} \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}^{37} u d u

Apply the Power Rule

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

Compute the boundaries:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{74 37}{3} - \frac{2}{3})

Simplify

= \frac{37 37 - 1}{12}

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What is the integral from 0 to 3 of sqrt(1+4r^2)r ?
The integral from 0 to 3 of sqrt(1+4r^2)r is (37sqrt(37)-1)/(12)