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

The integral from 1 to 4 of (x^2-4x-3) is -18

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

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

- 18

Solution steps

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

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 4 x d x = 30

\int_{1}^{4} 3 d x = 9

= 21 - 30 - 9

Simplify

= - 18

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What is the integral from 1 to 4 of (x^2-4x-3) ?
The integral from 1 to 4 of (x^2-4x-3) is -18