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

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

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integral from-1 to 1 of (1+x+x^2+x^3)

\int_{- 1}^{1} (1 + x + x^{2} + x^{3}) d x

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

\int_{- 1}^{1} 1 + x + x^{2} + x^{3} 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} x d x + \int_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} x^{3} d x

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

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

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

\int_{- 1}^{1} x^{3} d x = 0

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

Simplify

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

= \frac{8}{3}

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