What is the integral from 0 to pi of (2x-sin(x)) ?

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

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integral from 0 to pi of (2x-sin(x))

\int_{0}^{π} (2 x - sin (x)) d x

π^{2} - 2

Decimal Notation

7.86960 \dots

Solution steps

\int_{0}^{π} 2 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} 2 x d x - \int_{0}^{π} sin (x) d x

\int_{0}^{π} 2 x d x = π^{2}

\int_{0}^{π} sin (x) d x = 2

= π^{2} - 2

\int_{0}^{9} (9 - x) d x

\int_{0}^{x^{2}} x cos (y) d y

\int_{1}^{2} \frac{1}{1 + x ^{2}} d x

\int_{1}^{8} 5 d x

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

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