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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{2}^{4} t^{2} - 2 t - 3 d t

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} t^{2} d t - \int_{2}^{4} 2 t d t - \int_{2}^{4} 3 d t

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

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

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

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

Simplify

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

= \frac{2}{3}

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

\int_{1}^{e} x ln^{2} (x) d x

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

\int_{1}^{e} y ln (y) d y

\int_{0}^{x^{2}} \frac{1}{x ^{3} + 1} d x

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