What is the integral from 1 to-1 of (9x+8)^2 ?

The integral from 1 to-1 of (9x+8)^2 is -182

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integral from 1 to-1 of (9x+8)^2

\int_{1}^{- 1} (9 x + 8)^{2} d x

- 182

Solution steps

\int_{1}^{- 1} (9 x + 8)^{2} d x

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

= - \int_{- 1}^{1} (9 x + 8)^{2} d x

Expand

(9 x + 8)^{2} : 81 x^{2} + 144 x + 64

= - \int_{- 1}^{1} 81 x^{2} + 144 x + 64 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_{- 1}^{1} 81 x^{2} d x + \int_{- 1}^{1} 144 x d x + \int_{- 1}^{1} 64 d x)

\int_{- 1}^{1} 81 x^{2} d x = 54

\int_{- 1}^{1} 144 x d x = 0

\int_{- 1}^{1} 64 d x = 128

= - (54 + 0 + 128)

Simplify

= - 182

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What is the integral from 1 to-1 of (9x+8)^2 ?
The integral from 1 to-1 of (9x+8)^2 is -182