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

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

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integral from 0 to 1 of sin(2t)

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

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

Decimal Notation

0.70807 \dots

Solution steps

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

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)

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