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

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

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

\int_{0}^{π} (2 sin (x)) d x

4

Solution steps

\int_{0}^{π} 2 sin (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{π} sin (x) d x

Use the common integral:

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

= 2 [- cos (x)]_{0}^{π}

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

\int_{1}^{e} ln^{3} (x) d x

\int_{0}^{3} x e^{x^{2}} d x

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

\int_{1}^{9} \frac{u - 2 u ^{2}}{u} d u

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

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