What is the integral from 0 to 5 of t^3sqrt(9t^4+1) ?

The integral from 0 to 5 of t^3sqrt(9t^4+1) is (5626sqrt(5626)-1)/(54)

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integral from 0 to 5 of t^3sqrt(9t^4+1)

\int_{0}^{5} t^{3} 9 t^{4} + 1 d t

\frac{5626 5626 - 1}{54}

Decimal

7814.56490 \dots

Solution steps

\int_{0}^{5} t^{3} 9 t^{4} + 1 d t

Apply u-substitution

: \int_{1}^{5626} \frac{u}{36} d u

= \int_{1}^{5626} \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_{1}^{5626} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5626}

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5626}

Compute the boundaries:

\frac{11252 5626}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{11252 5626}{3} - \frac{2}{3})

Simplify

= \frac{5626 5626 - 1}{54}

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What is the integral from 0 to 5 of t^3sqrt(9t^4+1) ?
The integral from 0 to 5 of t^3sqrt(9t^4+1) is (5626sqrt(5626)-1)/(54)