What is the integral from 0 to 2 of (3-x)(1-x^2) ?

The integral from 0 to 2 of (3-x)(1-x^2) is 0

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integral from 0 to 2 of (3-x)(1-x^2)

\int_{0}^{2} (3 - x) (1 - x^{2}) d x

0

Solution steps

\int_{0}^{2} (3 - x) (1 - x^{2}) d x

Expand

(3 - x) (1 - x^{2}) : 3 - 3 x^{2} - x + x^{3}

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

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

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

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

\int_{0}^{2} x^{3} d x = 4

= 6 - 8 - 2 + 4

Simplify

= 0

What is the integral from 0 to 2 of (3-x)(1-x^2) ?
The integral from 0 to 2 of (3-x)(1-x^2) is 0