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

The integral from 0 to 2 of sin(2x) is 1/2 (-cos(4)+1)

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integral from 0 to 2 of sin(2x)

\int_{0}^{2} sin (2 x) d x

\frac{1}{2} (- cos (4) + 1)

Decimal

0.82682 \dots

Solution steps

\int_{0}^{2} sin (2 x) d x

Apply u-substitution

: \int_{0}^{4} sin (u) \frac{1}{2} d u

= \int_{0}^{4} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{4}

Compute the boundaries:

- cos (4) + 1

= \frac{1}{2} (- cos (4) + 1)

\int_{- 1}^{1} e^{y} d y

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

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

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

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

What is the integral from 0 to 2 of sin(2x) ?
The integral from 0 to 2 of sin(2x) is 1/2 (-cos(4)+1)