What is the integral from 0 to 1 of sin(2x) ?

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

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integral from 0 to 1 of sin(2x)

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

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

Decimal Notation

0.70807 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (2) + 1

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

x \to 0 lim (\frac{1}{ln ( 1 + x )} - \frac{1}{x})

\int 12 x sec^{2} (x) d x

\frac{\partial}{\partial y} ((x + 2 y^{3})^{\frac{1}{3}})

(x^{4} + 3 x)^{- 1}

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