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

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

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integral from 1 to 2 of sin(x)

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

- cos (2) + cos (1)

Decimal Notation

0.95644 \dots

Solution steps

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

Use the common integral:

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

= [- cos (x)]_{1}^{2}

Compute the boundaries:

- cos (2) + cos (1)

= - cos (2) + cos (1)

\frac{d}{d x} ((x^{2} - 4)^{2})

\frac{d y}{d x} = sin (3 x) - cos (5 x)

(3 t - 1) (4 t - 4)^{- 1}

\int x \frac{1}{2} d x

\frac{d}{d x} (2 x - 5)

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