What is the integral from 0 to pi of 1sin(x) ?

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

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integral from 0 to pi of 1sin(x)

\int_{0}^{π} 1 sin (x) d x

2

Solution steps

\int_{0}^{π} 1 \cdot sin (x) d x

Simplify

= \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

\int_{- \frac{π}{2}}^{\frac{π}{2}} 5 cos (t) d t

\int_{0}^{x} \frac{3}{( 1 + x ) ^{4}} d x

\int_{0}^{1} 4 1 - x^{2} d x

\int_{0}^{3} \frac{1}{x ^{2} - 8 x + 7} d x

\int_{- 3}^{3} (2 x + 6) d x

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