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

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

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integral from 0 to pi of-sin(x)

\int_{0}^{π} - sin (x) d x

- 2

Solution steps

\int_{0}^{π} - sin (x) d x

Take the constant out:

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

= - \int_{0}^{π} sin (x) d x

Use the common integral:

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

= - [- cos (x)]_{0}^{π}

Compute the boundaries:

2

= - 2

x \to 0 lim ((sin (x))^{t a n (x)})

f (x) = sin (sin (x))

\frac{d}{d x} (\frac{1}{4} x - \frac{1}{6})

\int \frac{8 x ^{3} - 16 x ^{2} + 5}{x ^{2} - 2 x} d x

\int e^{- s x} cos (x) d x

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