What is the integral from 0 to 1 of t^3sqrt(4+9t^4) ?

The integral from 0 to 1 of t^3sqrt(4+9t^4) is (13sqrt(13)-8)/(54)

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integral from 0 to 1 of t^3sqrt(4+9t^4)

\int_{0}^{1} t^{3} 4 + 9 t^{4} d t

\frac{13 13 - 8}{54}

Decimal

0.71985 \dots

Solution steps

\int_{0}^{1} t^{3} 4 + 9 t^{4} d t

Apply u-substitution

: \int_{4}^{13} \frac{u}{36} d u

= \int_{4}^{13} \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{4}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{16}{3}

= \frac{1}{36} (\frac{26 13}{3} - \frac{16}{3})

Simplify

= \frac{13 13 - 8}{54}

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What is the integral from 0 to 1 of t^3sqrt(4+9t^4) ?
The integral from 0 to 1 of t^3sqrt(4+9t^4) is (13sqrt(13)-8)/(54)