What is the integral from 2 to 3 of (8x^3+3x^2+6x) ?

The integral from 2 to 3 of (8x^3+3x^2+6x) is 164

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integral from 2 to 3 of (8x^3+3x^2+6x)

\int_{2}^{3} (8 x^{3} + 3 x^{2} + 6 x) d x

164

Solution steps

\int_{2}^{3} 8 x^{3} + 3 x^{2} + 6 x 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_{2}^{3} 8 x^{3} d x + \int_{2}^{3} 3 x^{2} d x + \int_{2}^{3} 6 x d x

\int_{2}^{3} 8 x^{3} d x = 130

\int_{2}^{3} 3 x^{2} d x = 19

\int_{2}^{3} 6 x d x = 15

= 130 + 19 + 15

Simplify

= 164

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What is the integral from 2 to 3 of (8x^3+3x^2+6x) ?
The integral from 2 to 3 of (8x^3+3x^2+6x) is 164