What is the integral from 1 to x^2 of 16xy^3 ?

The integral from 1 to x^2 of 16xy^3 is 4x(x^8-1)

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integral from 1 to x^2 of 16xy^3

\int_{1}^{x^{2}} 16 x y^{3} d y

4 x (x^{8} - 1)

Solution steps

\int_{1}^{x^{2}} 16 x y^{3} d y

Take the constant out:

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

= 16 x \cdot \int_{1}^{x^{2}} y^{3} d y

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{1}^{x^{2}}

= 16 x [\frac{y ^{4}}{4}]_{1}^{x^{2}}

Compute the boundaries:

\frac{x ^{8}}{4} - \frac{1}{4}

= 16 x (\frac{x ^{8}}{4} - \frac{1}{4})

Simplify

= 4 x (x^{8} - 1)

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What is the integral from 1 to x^2 of 16xy^3 ?
The integral from 1 to x^2 of 16xy^3 is 4x(x^8-1)