How do you find the average value of #f(x)=1/(x^2+1)# as x varies between #[-1,1]#?

Question

Christopher Agresta · Accepted Answer

Evaluate #1/(1-(-1)) int_-1^1 1/(x^2+1) dx = pi/4#

The average value of #f# on the interval #[a,b]# is

#1/(b-a) int_a^b f(x) dx#

We also need:

#int 1/(x^2+1) dx = tan^-1 x +C#

or, you may prefer the notation

#int 1/(x^2+1) dx = arctan x +C#.

The average value is

#1/(1-(-1)) int_-1^1 1/(x^2+1) dx =1/2 [tan^-1 x]_-1^1#

# = 1/2[pi/4 - (-pi/4)]#

# = pi/4#

Elijah Abram · Answer

undefined To find the average value of the function $ f(x) = \frac{1}{x^2 + 1} $ as $ x $ varies between $[-1, 1]$, we use the formula for the average value of a function on an interval $[a, b]$:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

In this case, $ a = -1 $ and $ b = 1 $. Substituting these values into the formula, we get:

$$ \text{Average value} = \frac{1}{1 - (-1)} \int_{-1}^{1} \frac{1}{x^2 + 1} \, dx $$

Now, we integrate $ \frac{1}{x^2 + 1} $ with respect to $ x $ from $-1$ to $1$:

$$ \int_{-1}^{1} \frac{1}{x^2 + 1} \, dx = \left[ \arctan(x) \right]_{-1}^{1} $$

Evaluate the antiderivative at the upper and lower limits:

$$ \arctan(1) - \arctan(-1) = \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) = \frac{\pi}{2} $$

So, the average value of $ f(x) = \frac{1}{x^2 + 1} $ as $ x $ varies between $[-1, 1]$ is $ \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4} $.