landdenaw

2022-06-22

How do you find the average value of $\mathrm{sin}x$ as x varies between $[0,\pi ]$?

Jaida Sanders

Beginner2022-06-23Added 18 answers

Explanation:

The average value of a function f on the interval [a,b] is found through the integral expression

$\frac{1}{b-a}{\int}_{1}^{b}f(x)dx$

Here, this gives us an average value of

$\frac{1}{\pi -0}{\int}_{0}^{\pi}\mathrm{sin}(x)dx$

The antiderivative of $\mathrm{sin}(x)$ is $-\mathrm{cos}(x):$

$=\frac{1}{\pi}[-\mathrm{cos}(x){]}_{0}^{\pi}$

$=\frac{1}{\pi}(-\mathrm{cos}(\pi )-(-\mathrm{cos}(0)))$

$=\frac{1}{\pi}(-(-1)-(-1))$

$=\frac{2}{\pi}$

The average value of a function f on the interval [a,b] is found through the integral expression

$\frac{1}{b-a}{\int}_{1}^{b}f(x)dx$

Here, this gives us an average value of

$\frac{1}{\pi -0}{\int}_{0}^{\pi}\mathrm{sin}(x)dx$

The antiderivative of $\mathrm{sin}(x)$ is $-\mathrm{cos}(x):$

$=\frac{1}{\pi}[-\mathrm{cos}(x){]}_{0}^{\pi}$

$=\frac{1}{\pi}(-\mathrm{cos}(\pi )-(-\mathrm{cos}(0)))$

$=\frac{1}{\pi}(-(-1)-(-1))$

$=\frac{2}{\pi}$