What is the integral from 5 to 8 of f(4x^3+1)x^2 ?

The integral from 5 to 8 of f(4x^3+1)x^2 is 164475f

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integral from 5 to 8 of f(4x^3+1)x^2

\int_{5}^{8} f (4 x^{3} + 1) x^{2} d x

164475 f

Solution steps

\int_{5}^{8} f (4 x^{3} + 1) 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_{5}^{8} (1 + 4 x^{3}) x^{2} d x

Expand

(1 + 4 x^{3}) x^{2} : x^{2} + 4 x^{5}

= f \cdot \int_{5}^{8} x^{2} + 4 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_{5}^{8} x^{2} d x + \int_{5}^{8} 4 x^{5} d x)

\int_{5}^{8} x^{2} d x = 129

\int_{5}^{8} 4 x^{5} d x = 164346

= f (129 + 164346)

Simplify

= 164475 f

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What is the integral from 5 to 8 of f(4x^3+1)x^2 ?
The integral from 5 to 8 of f(4x^3+1)x^2 is 164475f