What is the integral from 0 to 2 of (2x^3+3x^2-2x-3) ?

The integral from 0 to 2 of (2x^3+3x^2-2x-3) is 6

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integral from 0 to 2 of (2x^3+3x^2-2x-3)

\int_{0}^{2} (2 x^{3} + 3 x^{2} - 2 x - 3) d x

6

Solution steps

\int_{0}^{2} 2 x^{3} + 3 x^{2} - 2 x - 3 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_{0}^{2} 2 x^{3} d x + \int_{0}^{2} 3 x^{2} d x - \int_{0}^{2} 2 x d x - \int_{0}^{2} 3 d x

\int_{0}^{2} 2 x^{3} d x = 8

\int_{0}^{2} 3 x^{2} d x = 8

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} 3 d x = 6

= 8 + 8 - 4 - 6

Simplify

= 6

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What is the integral from 0 to 2 of (2x^3+3x^2-2x-3) ?
The integral from 0 to 2 of (2x^3+3x^2-2x-3) is 6