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

The integral from 0 to 2 of xf(x^2) is f*4

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

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

f \cdot 4

Solution steps

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

Simplify

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

= f \cdot \int_{0}^{2} x^{3} d x

Apply the Power Rule

: [\frac{x ^{4}}{4}]_{0}^{2}

= f [\frac{x ^{4}}{4}]_{0}^{2}

Compute the boundaries:

4

= f \cdot 4

\int_{- 2}^{2} x - \frac{4}{x} d x

\int_{- t}^{t} t d t

\int_{- 2}^{1} (- x + 2 - x^{2}) d x

\int_{0}^{3} (y - 3) (2 y + 1) d y

\int_{0}^{\infty} e^{- y^{2}} d y

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