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

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

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

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

\frac{4}{3}

Solution steps

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

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} x^{2} d x = \frac{26}{3}

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

= 16 - \frac{26}{3} - 6

Simplify

16 - \frac{26}{3} - 6 : \frac{4}{3}

= \frac{4}{3}

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