∫_0^π(1 sin8x)dx =∫_0^π1 dx ∫_0^πsin 8x dx =[x]_0^π [ cos 8x8]_0^π =π [cos 8x8]_π^0 =π [18 18] =π ...

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Definite Integral as Limit of Sum

Solve: ∫0π1-...

Question

Solve:
$\int_{0}^{π} (1 - sin 8 x) d x$

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Solution

$\int_{0}^{π} (1 - sin 8 x) d x$
$= \int_{0}^{π} 1 d x - \int_{0}^{π} sin 8 x d x$
$= [x]_{0}^{π} - [- \frac{cos 8 x}{8}]_{0}^{π}$
$= π - [\frac{cos 8 x}{8}]_{π}^{0}$
$= π - [\frac{1}{8} - \frac{1}{8}]$
$= π$

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