What is the integral from 3.3 to 8.3 of (12x^2-6x+4) ?

The integral from 3.3 to 8.3 of (12x^2-6x+4) is 1989.39999…

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integral from 3.3 to 8.3 of (12x^2-6x+4)

\int_{3.3}^{8.3} (12 x^{2} - 6 x + 4) d x

1989.39999 \dots

Solution steps

\int_{3.3}^{8.3} 12 x^{2} - 6 x + 4 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{3.3}^{8.3} 12 x^{2} d x - \int_{3.3}^{8.3} 6 x d x + \int_{3.3}^{8.3} 4 d x

\int_{3.3}^{8.3} 12 x^{2} d x = 2143.39999 \dots

\int_{3.3}^{8.3} 6 x d x = 174

\int_{3.3}^{8.3} 4 d x = 20

= 2143.39999 \dots - 174 + 20

Simplify

= 1989.39999 \dots

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What is the integral from 3.3 to 8.3 of (12x^2-6x+4) ?
The integral from 3.3 to 8.3 of (12x^2-6x+4) is 1989.39999…