What is the integral from 0 to 1 of (4-8v^3+16v^7) ?

The integral from 0 to 1 of (4-8v^3+16v^7) is 4

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integral from 0 to 1 of (4-8v^3+16v^7)

\int_{0}^{1} (4 - 8 v^{3} + 16 v^{7}) d v

4

Solution steps

\int_{0}^{1} 4 - 8 v^{3} + 16 v^{7} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 4 d v - \int_{0}^{1} 8 v^{3} d v + \int_{0}^{1} 16 v^{7} d v

\int_{0}^{1} 4 d v = 4

\int_{0}^{1} 8 v^{3} d v = 2

\int_{0}^{1} 16 v^{7} d v = 2

= 4 - 2 + 2

Simplify

= 4

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What is the integral from 0 to 1 of (4-8v^3+16v^7) ?
The integral from 0 to 1 of (4-8v^3+16v^7) is 4