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Numerade Educator



Problem 11 Easy Difficulty

(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) = 2 \sin x - \sin 2x $ , $ [0, \pi] $


(A). $f_{a v g}=\frac{4}{\pi}$
(B). $c \approx 1.24,2.81$


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Video Transcript

and this problem. We're using the technique of integration to find the average value of a function. And this comes based off of the mean value fear, um, that you might have learned in previous calculus courses or in calculus, one or the beginning of Cal. Gaby, if you're taking that. So let's review how we would find the average value of a function. So we stay on the closed interval a B that the average value of function F average is equivalent to one over B minus a times the integral of a to be of ffx in the X now in this case, were given. Ffx is to sign X minus sign of two x and r. Interval is from zero to pi and these air, including both of these endpoints. So now we can just plug in this information into the equation for the average value of the function. So we would get that the average value of the function is equivalent to one over pi minus zero times the integral from zero to pi off to sign next minus sign of two X with respect to X. So then we can just take the integral like how we would normally dio so we would find one over pi times the anti derivative of this function negative to cosign X plus one half times the co sign of two X and we're evaluating this from zero to pi. So then we would just plug in B minus a So we get one over pi times negative to co sign of pie plus one half times the coastline of two pi minus one over pi times negative to co side of zero plus one half times the cosine of zero. So now we would simplify all of this by looking at the points on the unit circle so we would get one over pi times two plus one, half minus one over pi times negative two plus one half. And then once you simplify this and get it all over pie, we would find that are the average value of our function is for over pie. So the next question comes from again the mean value. The're, um is what value of C would we find in this function or on this graph? So the way we would do that is we would set the average value of our function for over a pie equal to or a function. But instead of X, we replaced it with sees. So we would have for over pies equivalent to to sign C minus, sign of to see and we have to find C. Now, this function is tricky to do by hand, so we're going to use a graphical method. So what you would do is you would graph the function. Why? It was four over pie. So this is a very rough sketch of what that would look like. We'd have this curve and then our goal is to find the two points the Intersect, that line of why it was for over pie. And when we do that, we find that C is 1.24 and 2.81 And finally, we're told, let's graph the function that we just found the average value of. So this is what this would be our curve. This will be our interval right from zero to a little bit over three. And this dotted line represents the portion of our graph that we analyzed. So I hope this helps you understand how we can find the average value of a function, then how we find that value see in our function and lastly, how to graph it and understand visually what we calculated using integral calculus.