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

The integral from 0 to pi of 2sin(t) is 4

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integral from 0 to pi of 2sin(t)

\int_{0}^{π} 2 sin (t) d t

4

Solution steps

\int_{0}^{π} 2 sin (t) d t

Take the constant out:

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

= 2 \cdot \int_{0}^{π} sin (t) d t

Use the common integral:

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

= 2 [- cos (t)]_{0}^{π}

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

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

\int_{0}^{4 - z^{2}} z d z

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

\int_{- 3}^{1} (x^{2} - 2) - (2 x + 1) d x

\int_{2}^{5} 24 x^{2} d x

What is the integral from 0 to pi of 2sin(t) ?
The integral from 0 to pi of 2sin(t) is 4