What is the integral from 0 to t^4 of sqrt(u) ?

The integral from 0 to t^4 of sqrt(u) is (2t^6)/3

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integral from 0 to t^4 of sqrt(u)

\int_{0}^{t^{4}} u d u

\frac{2 t ^{6}}{3}

Solution steps

\int_{0}^{t^{4}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{t^{4}}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{t^{4}}

Compute the boundaries:

\frac{2}{3} t^{6}

= \frac{2}{3} t^{6}

Simplify

= \frac{2 t ^{6}}{3}

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What is the integral from 0 to t^4 of sqrt(u) ?
The integral from 0 to t^4 of sqrt(u) is (2t^6)/3