What is the integral from 0 to pi of (sin(x)-8x^3) ?

The integral from 0 to pi of (sin(x)-8x^3) is 2-2pi^4

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integral from 0 to pi of (sin(x)-8x^3)

\int_{0}^{π} (sin (x) - 8 x^{3}) d x

2 - 2 π^{4}

Solution steps

\int_{0}^{π} sin (x) - 8 x^{3} 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}^{π} sin (x) d x - \int_{0}^{π} 8 x^{3} d x

\int_{0}^{π} sin (x) d x = 2

\int_{0}^{π} 8 x^{3} d x = 2 π^{4}

= 2 - 2 π^{4}

\int_{x}^{1} 24 x (1 - y) d y

\int_{1}^{3} 4 x^{2} (x - 1) d x

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

\int_{s}^{s} 1 d x

\int_{0^{+}}^{1} \frac{1}{x ^{2}} d x

What is the integral from 0 to pi of (sin(x)-8x^3) ?
The integral from 0 to pi of (sin(x)-8x^3) is 2-2pi^4