What is the integral from 1 to 3 of 95+972t^2-18t ?

The integral from 1 to 3 of 95+972t^2-18t is 8542

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integral from 1 to 3 of 95+972t^2-18t

\int_{1}^{3} 95 + 972 t^{2} - 18 t d t

8542

Solution steps

\int_{1}^{3} 95 + 972 t^{2} - 18 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_{1}^{3} 95 d t + \int_{1}^{3} 972 t^{2} d t - \int_{1}^{3} 18 t d t

\int_{1}^{3} 95 d t = 190

\int_{1}^{3} 972 t^{2} d t = 8424

\int_{1}^{3} 18 t d t = 72

= 190 + 8424 - 72

Simplify

= 8542

What is the integral from 1 to 3 of 95+972t^2-18t ?
The integral from 1 to 3 of 95+972t^2-18t is 8542