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

The integral from 1 to 3 of (x^2-2x-4) is -22/3

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

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

- \frac{22}{3}

Decimal

- 7.33333 \dots

Solution steps

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

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

\int_{1}^{3} 2 x d x = 8

\int_{1}^{3} 4 d x = 8

= \frac{26}{3} - 8 - 8

Simplify

\frac{26}{3} - 8 - 8 : - \frac{22}{3}

= - \frac{22}{3}

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

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

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

\int_{0}^{x} x^{2} sin (x z) d z

\int_{0}^{s i n (x)} 7 t d t

What is the integral from 1 to 3 of (x^2-2x-4) ?
The integral from 1 to 3 of (x^2-2x-4) is -22/3