What is the integral from 0 to 4 of ((t^2)/3+2t) ?

The integral from 0 to 4 of ((t^2)/3+2t) is 208/9

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integral from 0 to 4 of ((t^2)/3+2t)

\int_{0}^{4} (\frac{t ^{2}}{3} + 2 t) d t

\frac{208}{9}

Decimal

23.11111 \dots

Solution steps

\int_{0}^{4} \frac{t ^{2}}{3} + 2 t 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_{0}^{4} \frac{t ^{2}}{3} d t + \int_{0}^{4} 2 t d t

\int_{0}^{4} \frac{t ^{2}}{3} d t = \frac{64}{9}

\int_{0}^{4} 2 t d t = 16

= \frac{64}{9} + 16

Simplify

\frac{64}{9} + 16 : \frac{208}{9}

= \frac{208}{9}

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What is the integral from 0 to 4 of ((t^2)/3+2t) ?
The integral from 0 to 4 of ((t^2)/3+2t) is 208/9