What is the integral from 0 to x^3 of sqrt(t+8) ?

The integral from 0 to x^3 of sqrt(t+8) is (2(x^3+8)^{3/2}-32sqrt(2))/3

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integral from 0 to x^3 of sqrt(t+8)

\int_{0}^{x^{3}} t + 8 d t

\frac{2 ( x ^{3} + 8 ) ^{\frac{3}{2}} - 32 2}{3}

Solution steps

\int_{0}^{x^{3}} t + 8 d t

Apply u-substitution

: \int_{8}^{x^{3} + 8} u d u

= \int_{8}^{x^{3} + 8} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{x^{3} + 8}

= [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{x^{3} + 8}

Compute the boundaries:

\frac{2}{3} (x^{3} + 8)^{\frac{3}{2}} - \frac{32 2}{3}

= \frac{2}{3} (x^{3} + 8)^{\frac{3}{2}} - \frac{32 2}{3}

Simplify

= \frac{2 ( x ^{3} + 8 ) ^{\frac{3}{2}} - 32 2}{3}

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What is the integral from 0 to x^3 of sqrt(t+8) ?
The integral from 0 to x^3 of sqrt(t+8) is (2(x^3+8)^{3/2}-32sqrt(2))/3