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

The integral from 1 to-1 of (r+1)^2 is -8/3

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

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

- \frac{8}{3}

Decimal

- 2.66666 \dots

Solution steps

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

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

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

Expand

(r + 1)^{2} : r^{2} + 2 r + 1

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

\int_{- 1}^{1} r^{2} d r = \frac{2}{3}

\int_{- 1}^{1} 2 r d r = 0

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

= - (\frac{2}{3} + 0 + 2)

Simplify

- (\frac{2}{3} + 0 + 2) : - \frac{8}{3}

= - \frac{8}{3}

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What is the integral from 1 to-1 of (r+1)^2 ?
The integral from 1 to-1 of (r+1)^2 is -8/3