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

The integral from 1 to 2 of 2x(x^2-1)^2 is 9

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

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

9

Solution steps

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

Take the constant out:

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

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

Expand

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

= 2 \cdot \int_{1}^{2} x^{5} - 2 x^{3} + x 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

= 2 (\int_{1}^{2} x^{5} d x - \int_{1}^{2} 2 x^{3} d x + \int_{1}^{2} x d x)

\int_{1}^{2} x^{5} d x = \frac{21}{2}

\int_{1}^{2} 2 x^{3} d x = \frac{15}{2}

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

= 2 (\frac{21}{2} - \frac{15}{2} + \frac{3}{2})

Simplify

2 (\frac{21}{2} - \frac{15}{2} + \frac{3}{2}) : 9

= 9

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What is the integral from 1 to 2 of 2x(x^2-1)^2 ?
The integral from 1 to 2 of 2x(x^2-1)^2 is 9