What is the integral from 0 to 4 of (2x+5)(3x-1) ?

The integral from 0 to 4 of (2x+5)(3x-1) is 212

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integral from 0 to 4 of (2x+5)(3x-1)

\int_{0}^{4} (2 x + 5) (3 x - 1) d x

212

Solution steps

\int_{0}^{4} (2 x + 5) (3 x - 1) d x

Expand

(2 x + 5) (3 x - 1) : 6 x^{2} + 13 x - 5

= \int_{0}^{4} 6 x^{2} + 13 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

= \int_{0}^{4} 6 x^{2} d x + \int_{0}^{4} 13 x d x - \int_{0}^{4} 5 d x

\int_{0}^{4} 6 x^{2} d x = 128

\int_{0}^{4} 13 x d x = 104

\int_{0}^{4} 5 d x = 20

= 128 + 104 - 20

Simplify

= 212

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What is the integral from 0 to 4 of (2x+5)(3x-1) ?
The integral from 0 to 4 of (2x+5)(3x-1) is 212