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

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

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

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

3

Solution steps

\int_{0}^{1} 3 - 8 x^{3} + 16 x^{7} 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_{0}^{1} 3 d x - \int_{0}^{1} 8 x^{3} d x + \int_{0}^{1} 16 x^{7} d x

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

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

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

= 3 - 2 + 2

Simplify

= 3

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