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

Decimal

5.14159 \dots

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

\frac{d y}{d t} = t \cdot y

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

\frac{d}{d x} (\frac{10 x + 4 x ^{2}}{( 5 + 4 x ) ^{2}})

\frac{d}{d x} ((2 x + 5)^{3} \cdot (3 x - 1)^{4})

\int (- 2^{3 - 2 x}) d x

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