What is the integral from x^2 to x^4 of sqrt(u) ?

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

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integral from x^2 to x^4 of sqrt(u)

\int_{x^{2}}^{x^{4}} u d u

\frac{2 x ^{6} - 2 x ^{3}}{3}

Solution steps

\int_{x^{2}}^{x^{4}} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3} x^{6} - \frac{2}{3} x^{3}

= \frac{2}{3} x^{6} - \frac{2}{3} x^{3}

Simplify

= \frac{2 x ^{6} - 2 x ^{3}}{3}

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