What is the integral from 0 to pi of 9sin(x)-sin(9x) ?

The integral from 0 to pi of 9sin(x)-sin(9x) is 160/9

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integral from 0 to pi of 9sin(x)-sin(9x)

\int_{0}^{π} 9 sin (x) - sin (9 x) d x

\frac{160}{9}

Decimal

17.77777 \dots

Solution steps

\int_{0}^{π} 9 sin (x) - sin (9 x) 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} 9 sin (x) d x - \int_{0}^{π} sin (9 x) d x

\int_{0}^{π} 9 sin (x) d x = 18

\int_{0}^{π} sin (9 x) d x = \frac{2}{9}

= 18 - \frac{2}{9}

Simplify

18 - \frac{2}{9} : \frac{160}{9}

= \frac{160}{9}

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\int_{- 3}^{3} 9 - x^{2} + 2 d x

\int_{- 1}^{3} (5 x + 1)^{5} d x

\int_{0}^{π} π d x

What is the integral from 0 to pi of 9sin(x)-sin(9x) ?
The integral from 0 to pi of 9sin(x)-sin(9x) is 160/9