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

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

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

\int_{0}^{2 π} 4 sin (t) d t

0

Solution steps

\int_{0}^{2 π} 4 sin (t) d t

Take the constant out:

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

= 4 \cdot \int_{0}^{2 π} sin (t) d t

Use the common integral:

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

= 4 [- cos (t)]_{0}^{2 π}

Compute the boundaries:

0

= 4 \cdot 0

Simplify

= 0

\int_{- 1}^{0} 3^{x - 1} d x

\int_{0}^{4} \frac{1}{2} (x)^{2} d x

\int_{0}^{8} 462 e^{0.019 t} d t

\int_{0}^{10} x e^{(1 - x^{2})} d x

\int_{1}^{2} 4 π (t - 2)^{2} d t

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