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

The integral from 0 to pi of (4sin(x)) is 8

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integral from 0 to pi of (4sin(x))

\int_{0}^{π} (4 sin (x)) d x

8

Solution steps

\int_{0}^{π} 4 sin (x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{π} sin (x) d x

Use the common integral:

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

= 4 [- cos (x)]_{0}^{π}

Compute the boundaries:

2

= 4 \cdot 2

Simplify

= 8

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

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

\int_{0}^{0.35} \frac{2}{x ^{2} - 4} d x

\int_{0}^{2} 9 x^{2} 1 + x^{3} d x

\int_{0}^{2 π} (1 + cos (x))^{2} d x

What is the integral from 0 to pi of (4sin(x)) ?
The integral from 0 to pi of (4sin(x)) is 8