What is the integral from 0 to pi of (1+sin(x)) ?

The integral from 0 to pi of (1+sin(x)) is pi+2

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integral from 0 to pi of (1+sin(x))

\int_{0}^{π} (1 + sin (x)) d x

π + 2

Solution steps

\int_{0}^{π} 1 + sin (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} 1 d x + \int_{0}^{π} sin (x) d x

\int_{0}^{π} 1 d x = π

\int_{0}^{π} sin (x) d x = 2

= π + 2

2 y^{^{'''}} + 3 y^{^{''}} - 23 y^{^{'}} - 12 y = 0

\int x sin (10 x) d x

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x \to 0 lim (x^{2} - 4 x + 2)

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What is the integral from 0 to pi of (1+sin(x)) ?
The integral from 0 to pi of (1+sin(x)) is pi+2