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

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

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

\int_{0}^{π} (x + sin (x)) d x

\frac{π ^{2}}{2} + 2

Decimal Notation

6.93480 \dots

Solution steps

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

\int_{0}^{π} x d x = \frac{π ^{2}}{2}

\int_{0}^{π} sin (x) d x = 2

= \frac{π ^{2}}{2} + 2

a^{5} + cos^{5} (x)

n = 1 \sum \infty (- 3)^{- 3 n - 1}

f (x) = e^{3 x}, (0, 1)

\frac{d}{d x} ((2 x + 5) (x + 1))

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

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