What is the integral from 0 to pi of (8+sin(x)) ?

The integral from 0 to pi of (8+sin(x)) is 8pi+2

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integral from 0 to pi of (8+sin(x))

\int_{0}^{π} (8 + sin (x)) d x

8 π + 2

Decimal Notation

27.13274 \dots

Solution steps

\int_{0}^{π} 8 + 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} 8 d x + \int_{0}^{π} sin (x) d x

\int_{0}^{π} 8 d x = 8 π

\int_{0}^{π} sin (x) d x = 2

= 8 π + 2

\int_{1}^{2} sin (π t) d t

\int_{1}^{\infty} \frac{5}{x ^{2} + 4} d x

\int_{0}^{1} 1 - u^{2} d u

\int_{- 2}^{8} (8 x + 6) d x

\int_{0}^{π} cos (x) (x) d x

What is the integral from 0 to pi of (8+sin(x)) ?
The integral from 0 to pi of (8+sin(x)) is 8pi+2