What is the integral from-1 to 1 of (x^7+1)(3x^2+1) ?

The integral from-1 to 1 of (x^7+1)(3x^2+1) is 4

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integral from-1 to 1 of (x^7+1)(3x^2+1)

\int_{- 1}^{1} (x^{7} + 1) (3 x^{2} + 1) d x

4

Solution steps

\int_{- 1}^{1} (x^{7} + 1) (3 x^{2} + 1) d x

Expand

(x^{7} + 1) (3 x^{2} + 1) : 3 x^{9} + x^{7} + 3 x^{2} + 1

= \int_{- 1}^{1} 3 x^{9} + x^{7} + 3 x^{2} + 1 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} 3 x^{9} d x + \int_{- 1}^{1} x^{7} d x + \int_{- 1}^{1} 3 x^{2} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 3 x^{9} d x = 0

\int_{- 1}^{1} x^{7} d x = 0

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

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

= 0 + 0 + 2 + 2

Simplify

= 4

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What is the integral from-1 to 1 of (x^7+1)(3x^2+1) ?
The integral from-1 to 1 of (x^7+1)(3x^2+1) is 4