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

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