What is the integral from 3 to x of (t^2+t+1) ?

The integral from 3 to x of (t^2+t+1) is x+(x^3)/3+(x^2-9)/2-12

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integral from 3 to x of (t^2+t+1)

\int_{3}^{x} (t^{2} + t + 1) d t

x + \frac{x ^{3}}{3} + \frac{x ^{2} - 9}{2} - 12

Solution steps

\int_{3}^{x} t^{2} + t + 1 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_{3}^{x} t^{2} d t + \int_{3}^{x} t d t + \int_{3}^{x} 1 d t

\int_{3}^{x} t^{2} d t = \frac{x ^{3}}{3} - 9

\int_{3}^{x} t d t = \frac{x ^{2} - 9}{2}

\int_{3}^{x} 1 d t = x - 3

= \frac{x ^{3}}{3} - 9 + \frac{x ^{2} - 9}{2} + x - 3

Simplify

= x + \frac{x ^{3}}{3} + \frac{x ^{2} - 9}{2} - 12

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What is the integral from 3 to x of (t^2+t+1) ?
The integral from 3 to x of (t^2+t+1) is x+(x^3)/3+(x^2-9)/2-12