# How do you find the average value of #f(x)=x/sqrt(x^2+1), 0<=x<=4#?

The average value of

This is a known integral.

Evaluate this using the 2nd fundamental theorem of calculus.

Hopefully this helps!

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Alternative to get

Note that the limits of integration changed because we were doing a u substitution. 0 became 1 because:

Similarly plugging in 4 became 17 because:

Now rewrite it so it's easier to integrate by treating the square root in the denominator:

This integral becomes

Take the 2 out and plug in 17 and 1. As you can see, this will give you the same answer as HSBC244's answer but does not require a trig substitution.

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To find the average value of ( f(x) = \frac{x}{\sqrt{x^2 + 1}} ) over the interval ( 0 \leq x \leq 4 ), follow these steps:

- Compute the definite integral of ( f(x) ) over the given interval.
- Divide the result from step 1 by the length of the interval.

The average value of ( f(x) ) over the interval ( 0 \leq x \leq 4 ) is approximately 1.3333.

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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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