What is the integral from 1 to 3 of 2x(x^2+6) ?

The integral from 1 to 3 of 2x(x^2+6) is 88

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integral from 1 to 3 of 2x(x^2+6)

\int_{1}^{3} 2 x (x^{2} + 6) d x

88

Solution steps

\int_{1}^{3} 2 x (x^{2} + 6) d x

Take the constant out:

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

= 2 \cdot \int_{1}^{3} x (x^{2} + 6) d x

Expand

x (x^{2} + 6) : x^{3} + 6 x

= 2 \cdot \int_{1}^{3} x^{3} + 6 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}^{3} x^{3} d x + \int_{1}^{3} 6 x d x)

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

\int_{1}^{3} 6 x d x = 24

= 2 (20 + 24)

Simplify

= 88

What is the integral from 1 to 3 of 2x(x^2+6) ?
The integral from 1 to 3 of 2x(x^2+6) is 88