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

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

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integral from 2 to 1 of sin(x)

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

cos (2) - cos (1)

Decimal

- 0.95644 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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))

Simplify

= cos (2) - cos (1)

\int_{0}^{2 x + y} 1 d z

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

\int_{0}^{10} t + 4 d t

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

\int_{1}^{\infty} \frac{1}{2 x + 2} d x

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