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

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

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

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

\frac{2}{3}

Solution steps

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

\int_{2}^{4} x^{2} d x = \frac{56}{3}

\int_{2}^{4} 2 x d x = 12

\int_{2}^{4} 3 d x = 6

= \frac{56}{3} - 12 - 6

Simplify

\frac{56}{3} - 12 - 6 : \frac{2}{3}

= \frac{2}{3}

\int_{0}^{1} x 3 - 2 x^{2} d x

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

\int_{0}^{\frac{π}{2}} 3 sin (y) cos (y) d y

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

\int_{0}^{1} 1 + 2 x d x

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