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Popular Calculus >

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

Solution

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

Solution

- 182

Solution steps

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

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

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

Expand

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

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

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

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

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

= - (54 + 0 + 128)

Simplify

= - 182

Graph

Popular Examples

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integral from 0 to 4 of ln(x)

integral from 0 to 1-x of 2

integral from 0 to 1 of 1/((1+x^2))

integral from 0 to 8 of 1/(x^2+64)

Frequently Asked Questions (FAQ)

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