What is the integral from 0 to t^6 of sqrt(u^7) ?

The integral from 0 to t^6 of sqrt(u^7) is (2t^{27})/9

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integral from 0 to t^6 of sqrt(u^7)

\int_{0}^{t^{6}} u^{7} d u

\frac{2 t ^{27}}{9}

Solution steps

\int_{0}^{t^{6}} u^{7} d u

u^{7} = u^{\frac{7}{2}},

assuming

u \geq 0

= \int_{0}^{t^{6}} u^{\frac{7}{2}} d u

Apply the Power Rule

: [\frac{2}{9} u^{\frac{9}{2}}]_{0}^{t^{6}}

= [\frac{2}{9} u^{\frac{9}{2}}]_{0}^{t^{6}}

Compute the boundaries:

\frac{2}{9} t^{27}

= \frac{2}{9} t^{27}

Simplify

= \frac{2 t ^{27}}{9}

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What is the integral from 0 to t^6 of sqrt(u^7) ?
The integral from 0 to t^6 of sqrt(u^7) is (2t^{27})/9