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

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

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

\int_{- π}^{0} (π + x) d x

π^{2} - \frac{π ^{2}}{2}

Decimal

4.93480 \dots

Solution steps

\int_{- π}^{0} π + 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_{- π}^{0} π d x + \int_{- π}^{0} x d x

\int_{- π}^{0} π d x = π^{2}

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

= π^{2} - \frac{π ^{2}}{2}

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