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

The integral from 1 to 3 of (3x^2-1)(6x) is 336

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

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

336

Solution steps

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

Take the constant out:

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

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

Expand

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

= 6 \cdot \int_{1}^{3} 3 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

= 6 (\int_{1}^{3} 3 x^{3} d x - \int_{1}^{3} x d x)

\int_{1}^{3} 3 x^{3} d x = 60

\int_{1}^{3} x d x = 4

= 6 (60 - 4)

Simplify

= 336

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What is the integral from 1 to 3 of (3x^2-1)(6x) ?
The integral from 1 to 3 of (3x^2-1)(6x) is 336