What is the integral from 5 to 4 of sin(t) ?

The integral from 5 to 4 of sin(t) is cos(5)-cos(4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 5 to 4 of sin(t)

\int_{5}^{4} sin (t) d t

cos (5) - cos (4)

Decimal

0.93730 \dots

Solution steps

\int_{5}^{4} sin (t) d t

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

= - \int_{4}^{5} sin (t) d t

Use the common integral:

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

= - [- cos (t)]_{4}^{5}

Compute the boundaries:

- cos (5) + cos (4)

= - (- cos (5) + cos (4))

Simplify

= cos (5) - cos (4)

\int_{0}^{5} (5 - ∣ x - 5 ∣) d x

\int_{1}^{6} \frac{5}{x} d x

\int_{0}^{1} k x d x

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

\int_{3}^{x} t^{2} + 1 d t

What is the integral from 5 to 4 of sin(t) ?
The integral from 5 to 4 of sin(t) is cos(5)-cos(4)