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

The integral from 0 to pi/2 of (2x+sin(x)) is (pi^2)/4+1

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

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

\frac{π ^{2}}{4} + 1

Decimal

3.46740 \dots

Solution steps

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

\int_{0}^{\frac{π}{2}} 2 x d x = \frac{π ^{2}}{4}

\int_{0}^{\frac{π}{2}} sin (x) d x = 1

= \frac{π ^{2}}{4} + 1

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What is the integral from 0 to pi/2 of (2x+sin(x)) ?
The integral from 0 to pi/2 of (2x+sin(x)) is (pi^2)/4+1