What is the integral from 0 to 9 of xf(x^2) ?

The integral from 0 to 9 of xf(x^2) is (6561f)/4

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integral from 0 to 9 of xf(x^2)

\int_{0}^{9} x f (x^{2}) d x

\frac{6561 f}{4}

Solution steps

\int_{0}^{9} x f x^{2} d x

Take the constant out:

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

= f \cdot \int_{0}^{9} x x^{2} d x

Simplify

x x^{2} : x^{3}

= f \cdot \int_{0}^{9} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{9}

= f [\frac{x ^{4}}{4}]_{0}^{9}

Compute the boundaries:

\frac{6561}{4}

= f \frac{6561}{4}

Simplify

= \frac{6561 f}{4}

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What is the integral from 0 to 9 of xf(x^2) ?
The integral from 0 to 9 of xf(x^2) is (6561f)/4