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

The integral from 1 to 3 of 4x^2(x-1) is 136/3

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integral from 1 to 3 of 4x^2(x-1)

\int_{1}^{3} 4 x^{2} (x - 1) d x

\frac{136}{3}

Solution steps

\int_{1}^{3} 4 x^{2} (x - 1) d x

Take the constant out:

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

= 4 \cdot \int_{1}^{3} x^{2} (x - 1) d x

Expand

x^{2} (x - 1) : x^{3} - x^{2}

= 4 \cdot \int_{1}^{3} x^{3} - x^{2} 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

= 4 (\int_{1}^{3} x^{3} d x - \int_{1}^{3} x^{2} d x)

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} x^{2} d x = \frac{26}{3}

= 4 (20 - \frac{26}{3})

Simplify

4 (20 - \frac{26}{3}) : \frac{136}{3}

= \frac{136}{3}

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What is the integral from 1 to 3 of 4x^2(x-1) ?
The integral from 1 to 3 of 4x^2(x-1) is 136/3