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

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

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integral from 0 to pi of sin(z)

\int_{0}^{π} sin (z) d z

2

Solution steps

\int_{0}^{π} sin (z) d z

Use the common integral:

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

= [- cos (z)]_{0}^{π}

Compute the boundaries:

2

= 2

\int_{1}^{e} \frac{( x ^{2} + 1 ) ln ( x )}{x} d x

\int_{0}^{2} 0.5 d x

\int_{0}^{1} x^{2} 6 x + 9 d x

\int_{0}^{80} (1 - \frac{t}{80})^{a} d t

\int_{0}^{2 π} \frac{1 + cos ( 2 θ )}{2} d θ

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