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

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

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

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

4

Solution steps

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

Expand

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

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

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

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

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

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

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

= 0 + 0 + 2 + 2

Simplify

= 4

\int_{0}^{4} x \cdot (3 - 5 x) d x

\int_{1}^{4} x e^{2 x} d x

\int_{1}^{2} \frac{3}{7} x^{2} d x

\int_{- 1}^{6} 6 d x

\int_{0}^{1} x^{2} e^{- 11 x} d x

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