What is the integral from 0 to 2 of x^2f(x^3+1) ?

The integral from 0 to 2 of x^2f(x^3+1) is 40/3 f

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integral from 0 to 2 of x^2f(x^3+1)

\int_{0}^{2} x^{2} f (x^{3} + 1) d x

\frac{40}{3} f

Solution steps

\int_{0}^{2} x^{2} f (x^{3} + 1) 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} (1 + x^{3}) x^{2} d x

Expand

(1 + x^{3}) x^{2} : x^{2} + x^{5}

= f \cdot \int_{0}^{2} x^{2} + x^{5} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= f (\int_{0}^{2} x^{2} d x + \int_{0}^{2} x^{5} d x)

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} x^{5} d x = \frac{32}{3}

= f (\frac{8}{3} + \frac{32}{3})

Join

\frac{8}{3} + \frac{32}{3} : \frac{40}{3}

= \frac{40}{3} f

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What is the integral from 0 to 2 of x^2f(x^3+1) ?
The integral from 0 to 2 of x^2f(x^3+1) is 40/3 f