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

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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} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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