What is the integral from 4 to-4 of (3r+2)^2 ?

The integral from 4 to-4 of (3r+2)^2 is -416

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integral from 4 to-4 of (3r+2)^2

\int_{4}^{- 4} (3 r + 2)^{2} d r

- 416

Solution steps

\int_{4}^{- 4} (3 r + 2)^{2} d r

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

= - \int_{- 4}^{4} (3 r + 2)^{2} d r

Expand

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

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

\int_{- 4}^{4} 9 r^{2} d r = 384

\int_{- 4}^{4} 12 r d r = 0

\int_{- 4}^{4} 4 d r = 32

= - (384 + 0 + 32)

Simplify

= - 416

\int_{- 2}^{2} 8 - x^{2} - x^{2} d x

\int_{- 6}^{4} 24 - 2 x - x^{2} d x

\int_{0}^{t} (y (u) - 1) d u

\int_{0}^{2} 12 x^{3} x^{4} + 9 d x

\int_{0}^{1} x^{2} x + 1 d x

What is the integral from 4 to-4 of (3r+2)^2 ?
The integral from 4 to-4 of (3r+2)^2 is -416