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

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

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

\int_{- π}^{0} sin (x) + x d x

- 2 - \frac{π ^{2}}{2}

Decimal Notation

- 6.93480 \dots

Solution steps

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

\int_{- π}^{0} sin (x) d x = - 2

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

= - 2 - \frac{π ^{2}}{2}

\frac{d}{d x} (e^{6 x} cos (b x))

(x + y)^{2} - 4 x + y + 10

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

\int \frac{2}{7} x^{2} + 1 d x

f (x) = 3 x^{2}, (- 1, 3)

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