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

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

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

\int_{- π}^{0} - sin (x) d x

2

Solution steps

\int_{- π}^{0} - sin (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{- π}^{0} sin (x) d x

Use the common integral:

\int sin (x) d x = - cos (x)

= - [- cos (x)]_{- π}^{0}

Compute the boundaries:

- 2

= - (- 2)

Simplify

= 2

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

\int_{0}^{2} - x^{2} - 3 x d x

\int_{0}^{4} y^{2} d y

\int_{0}^{2 π} - 16 sin^{2} (t) d t

\int_{1}^{3} x 10 x^{2} - 9 d x

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