What is the integral from 0 to 3 of 200+10t+1/3 t^2 ?

The integral from 0 to 3 of 200+10t+1/3 t^2 is 648

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integral from 0 to 3 of 200+10t+1/3 t^2

\int_{0}^{3} 200 + 10 t + \frac{1}{3} t^{2} d t

648

Solution steps

\int_{0}^{3} 200 + 10 t + \frac{1}{3} t^{2} 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}^{3} 200 d t + \int_{0}^{3} 10 t d t + \int_{0}^{3} \frac{1}{3} t^{2} d t

\int_{0}^{3} 200 d t = 600

\int_{0}^{3} 10 t d t = 45

\int_{0}^{3} \frac{1}{3} t^{2} d t = 3

= 600 + 45 + 3

Simplify

= 648

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What is the integral from 0 to 3 of 200+10t+1/3 t^2 ?
The integral from 0 to 3 of 200+10t+1/3 t^2 is 648