What is the integral from 0 to x^2 of sqrt(4x^2+5y) ?

The integral from 0 to x^2 of sqrt(4x^2+5y) is (38x^3)/(15)

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integral from 0 to x^2 of sqrt(4x^2+5y)

\int_{0}^{x^{2}} 4 x^{2} + 5 y d y

\frac{38 x ^{3}}{15}

Solution steps

\int_{0}^{x^{2}} 4 x^{2} + 5 y d y

Apply u-substitution

: \int_{4 x^{2}}^{9 x^{2}} \frac{u}{5} d u

= \int_{4 x^{2}}^{9 x^{2}} \frac{u}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{4 x^{2}}^{9 x^{2}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4 x^{2}}^{9 x^{2}}

= \frac{1}{5} [\frac{2}{3} u^{\frac{3}{2}}]_{4 x^{2}}^{9 x^{2}}

Compute the boundaries:

\frac{38}{3} x^{3}

= \frac{1}{5} \cdot \frac{38}{3} x^{3}

Simplify

= \frac{38 x ^{3}}{15}

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What is the integral from 0 to x^2 of sqrt(4x^2+5y) ?
The integral from 0 to x^2 of sqrt(4x^2+5y) is (38x^3)/(15)