What is the integral from 2 to 5 of 2000t+1000 ?

The integral from 2 to 5 of 2000t+1000 is 24000

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integral from 2 to 5 of 2000t+1000

\int_{2}^{5} 2000 t + 1000 d t

24000

Solution steps

\int_{2}^{5} 2000 t + 1000 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}^{5} 2000 t d t + \int_{2}^{5} 1000 d t

\int_{2}^{5} 2000 t d t = 21000

\int_{2}^{5} 1000 d t = 3000

= 21000 + 3000

Simplify

= 24000

\int_{0}^{20} 5 x d x

\int_{1}^{x} x^{5} e^{- 2 x} d x

\int_{0}^{0.6} 2 e^{- 1.5 x} d x

\int_{- 2}^{2} ∣ x ∣ - x^{2} + 2 d x

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

What is the integral from 2 to 5 of 2000t+1000 ?
The integral from 2 to 5 of 2000t+1000 is 24000