What is the integral from 4 to 3 of sin(t) ?

The integral from 4 to 3 of sin(t) is cos(4)-cos(3)

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integral from 4 to 3 of sin(t)

\int_{4}^{3} sin (t) d t

cos (4) - cos (3)

Solution steps

\int_{4}^{3} sin (t) d t

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

= - \int_{3}^{4} sin (t) d t

Use the common integral:

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

= - [- cos (t)]_{3}^{4}

Compute the boundaries:

- cos (4) + cos (3)

= - (- cos (4) + cos (3))

Simplify

= cos (4) - cos (3)

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What is the integral from 4 to 3 of sin(t) ?
The integral from 4 to 3 of sin(t) is cos(4)-cos(3)