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

The integral from 0 to pi/2 of-sin(x) is -1

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

\int_{0}^{\frac{π}{2}} - sin (x) d x

- 1

Solution steps

\int_{0}^{\frac{π}{2}} - sin (x) d x

Take the constant out:

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

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

Use the common integral:

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

= - [- cos (x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= - 1

\int_{1}^{5} \frac{2}{( x - 3 ) ^{4}} d x

\int_{0}^{\frac{π}{2}} 3 sin (\frac{x}{2}) d x

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

\int_{0}^{1} 2 + x^{2} d x

\int_{0}^{4 π} sin (x) d x

What is the integral from 0 to pi/2 of-sin(x) ?
The integral from 0 to pi/2 of-sin(x) is -1