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(a) Find the average value of $ f $ on the given interval.(b) Find $ c $ such that $ f_{ave} = f(c) $.(c) Sketch the graph of $ f $ and a rectangle whose area is the same as the area under the graph of $ f $.

$ f(x) = 2xe^{-x^2} $ , $ [0, 2] $

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(A). $f_{\text { ave }}=\frac{1-e^{-4}}{2}$(B). $2 c e^{-c^{2}}=\frac{1-e^{-4}}{2}$(C). Click to see the graph

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 5

Average Value of a Function

Missouri State University

Harvey Mudd College

Idaho State University

Lectures

01:01

(a) Find the average value…

06:23

03:11

03:26

04:23

03:52

02:47

06:29

01:07

03:08

03:42

Yeah. And this problem. We are finding the average value of a function using the technique of integration. And this is very similar in the stems from the mean value fear, um, which you might have learned previously in Cal Kwan born KLKB s O. This is a direct application using integration. So let's first understand how we would find the average value of a function on the closed interval A B. The average value of our function is equivalent to one over B minus a time to the integral from A to B of F of X in D X. Now, in this case for this problem, we're told that ffx is equivalent to two x e race negative X squared on the close intervals 0 to 2. So now we can plug all this information into the formula for the average value of a function. So we would get that the average value of our function would be equivalent to one over to minus zero times the integral from 0 to 2 of two x e raised the negative x squared with respect to X. Now we can use U substitution to make this a little bit easier to solve. So we're going to say Let u equal Negative X squared. So do you would be equal to negative two x in D X And then we can also change our limits of integration So we'll have our limits of integration. Being 0 to 2 will apply our substitution to you so we'll take negative zero squared with negative two squared, and we'll get our new limits as zero to negative four. So then we can continue with our integral will get that the average value of our function is now equal to one half times the integral from zero negative four of the negative e raised to the U with respect to you. And now we can take the anti derivative and we'll find out this is equivalent to 1/2 times negative. Either the you evaluated from 0 to 2. So then we're going to plug in our limits of integration. Using B minus A, we'll get one half times negative e race, the negative fourth, plus the race to the zero power. So when we simplify this and put it all over to will get that the average value of our function is one minus erase the negative. Fourth over to and now on the part B, we're told we have to find this value See, based on the mean value theorem of Integral, that this function is satisfied. So how we do this is we set our function where now see, is a substitute for X equal to the average value of our function. So we'll have to see raise t e Pardon me to see times e raised to negative C squared is equivalent to one minus e to the negative Fourth over to now. When you look at this, we have to find C. And that looks a little tricky to dio using algebraic techniques. So we're going to use a graphical method. So all we do is we would graph our function and we're going to also graph the line y equals. Why minus e erase the negative fourth over to and find the intersections on that curve. And when we do, we find that C is equivalent to 0.263 and 1.287 And now finally, we have to graph the function that we just found the average value for and this is what our function will look like. This is a rough sketch. But remember, our interval was from 0 to 2 and we found this portion in the these dotted lines. That's the portion that we analyzed using integral calculus. So I hope that this problem showed you how we confined the average value of a function using integral calculus and the mean value theorem for integral. And then how we find the value, See that equals or the average value of our function and how we can interpret that visually.

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