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

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

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

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

\frac{8}{3}

Solution steps

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

Expand

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

= \int_{- 1}^{1} 1 + 2 x + x^{2} 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} 1 d x + \int_{- 1}^{1} 2 x d x + \int_{- 1}^{1} x^{2} d x

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

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

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

= 2 + 0 + \frac{2}{3}

Simplify

2 + 0 + \frac{2}{3} : \frac{8}{3}

= \frac{8}{3}

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