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

The integral from 2 to-2 of (3r+2)^2 is -64

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integral from 2 to-2 of (3r+2)^2

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

- 64

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{2} (3 r + 2)^{2} d r

Expand

(3 r + 2)^{2} : 9 r^{2} + 12 r + 4

= - \int_{- 2}^{2} 9 r^{2} + 12 r + 4 d r

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}^{2} 9 r^{2} d r + \int_{- 2}^{2} 12 r d r + \int_{- 2}^{2} 4 d r)

\int_{- 2}^{2} 9 r^{2} d r = 48

\int_{- 2}^{2} 12 r d r = 0

\int_{- 2}^{2} 4 d r = 16

= - (48 + 0 + 16)

Simplify

= - 64

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What is the integral from 2 to-2 of (3r+2)^2 ?
The integral from 2 to-2 of (3r+2)^2 is -64