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

The integral from 0 to pi/2 of-xcos(x)+x is -pi/2+(pi^2)/8+1

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

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

- \frac{π}{2} + \frac{π ^{2}}{8} + 1

Decimal

0.66290 \dots

Solution steps

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

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

\int_{0}^{\frac{π}{2}} x d x = \frac{π ^{2}}{8}

= - (\frac{π}{2} - 1) + \frac{π ^{2}}{8}

Simplify

= - \frac{π}{2} + \frac{π ^{2}}{8} + 1

What is the integral from 0 to pi/2 of-xcos(x)+x ?
The integral from 0 to pi/2 of-xcos(x)+x is -pi/2+(pi^2)/8+1