Calculus 2 : Average Values and Lengths of Functions

Study concepts, example questions & explanations for Calculus 2

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Example Questions

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Example Question #31 : Average Values And Lengths Of Functions

Find the interval  over which the average value of the function  is equal to  (in other words find ). Note that 




Possible Answers:

Correct answer:


Find the interval  over which the average value of the function  defined below is equal to  (in other words find ). 



Definition of the average value of a function:

The average value of a function  over an interval  is defined as, 




We are given the left-hand end point  of the interval, and must find the right-hand endpoint  such that the average value works out to 








There are two real solutions for , we want the positive solution, therefore 










Example Question #32 : Average Values And Lengths Of Functions

You will need a scientific calculator for this problem.

Find the arc length of the function, , from to . (Round to the nearest hundredth)

Possible Answers:

Correct answer:


The formula for arc length for a function of x is

,where "S" is the arc length of f(x) from x=a to x=b.

Thus we first need to find . Using the power rule on each term gives,

Now substitute this into the arc length equation. Also put the bounds on the integral, x=0 and x=1, in for "a" and "b".

While it may be tempting to multiply out the , it is not necessary. The only way to integrate this arrangement is through a Trigonometric Substitution.

For this problem, we make the following substitution.

This will create a pythagorean trig identity inside the square root. But we need to find a few other parts using this substitution. Since we are changing the variable from to , we have to change into an equivalent expression in terms of .

Solve the substition for x.

Now differentiate both sides.

This is what we will substitute in for .

Our bounds are x=0 and x=1, but now we are integrating with respect to the new variable, . Thus we have find equivalent bounds. We do this by using the substitution we made and plugging in the x bounds, then solving for .

Start with

For x=0, replace x with 0 and solve for .

Repeat for x=1

Assembling all these pieces gives the following substitution:

Now apply the Pythagorean Trig. Identity, .

Simplify by canceling the square and square root. Then combine the secants through exponent properties.

The only way to integrate is to use integration by parts, and solving for the integral. First we can pull the 2 in the denominator out of the integral as a constant. 

Then we separate the into  and 

For now we will focus on just the integral, and will multiply by 1/2 afterward. Now use integration by parts with the following "u" and "dv".


then differentiate "u" to find "du" and integrate "dv" to find "v".


Now assemble the pieces. Remember that we are leaving the 1/2 off until the end.

Now combine the two  in the integral into . Also, combine the into .

Now apply a Pythagorean Trig Identity to the to make it .

Now mulitiply the   in the integral. 

Now split the integral into two integrals. Remember to distribute the integral's negative sign to both new integrals.

Notice that we have on both sides of the equation now. Thus we can solve for it algebraicly. Add it to both sides, so it cancels off the right side and then combine it on the left.

Before we divide both sides by 2, we should go ahead an evalute the integral on the right, and plug in the bounds for both terms. This will get a constant that will be easier to work with. (The following work only shows the right side of the equation.)

First, evaluate the

Then we will plug in the upper and lower bounds for each part. This is where a scientific calculator is required.

Plugging all this into a calculator gives

Now let's introduce this number in the integration by parts equation.

Now divide by the 2.

Now we plug this back into the Arc Length formula.

Evaluating this gives us our answer, rounded to the hundredths digit.

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