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

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

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

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

7

Solution steps

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

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

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

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

= 7 - 2 + 2

Simplify

= 7

\int_{0}^{7} x^{2} - 4 d x

\int_{0}^{1} (x^{3} + x^{2}) d x

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

\int_{0}^{4} 8 x - 32 d x

\int_{- π}^{π} - sin (x) d x

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