<p>Does Average Value Theorem work when the function is not continuous? If so... this Barron's book is crazy -_-</p>
<p>According to my calculus teacher, not necessarily.
Like the mean value theorem, which requires continuity and differentiability to ENSURE a value c such that f’(c)=average slope, the average value theorem requires continuity to ENSURE a c value such that f(c)=b-a*integral from a to b of f(x) dx. So while that c value may still exist, because the function is not continuous you can’t use average value theorem to prove it because the discontinuity could result in a scenario where there is no average value in the interval. I hope that made sense, lol.</p>
<p>Well the question asked me to find the average value, and gave a graph with a jump discontinuity… I circled “Cannot be found because it is not continuous” but in the answers they actually did the average value theorem.</p>
<p>I think that you could try finding an average value and if that value is actually a point on the graph then it could work. Don’t quote me on that though.</p>
<p>I have another question:</p>
<p>If region R is the base of a solid, and the cross sections perpendicular to the x axis are semicircles, find the volume of the solid.</p>
<p>Why is this the answer:</p>
<p>pi/2 (integrate from x to y)((1/2) Radius)^2?</p>
<p>Why is there an extra 0.5 inside of the parentheses? I hope this question is clear :/</p>
<p>I think by radius they mean top curve-bottom curve. If that is the case, by Radius they mean the diameter of the semicircle so it must be divided by 2 to find the radius.</p>
<p>find avg value on each interval and avg those w/a weighted avg</p>
<p>You can also think of the average value as the height of a rectangle that has both an equal area as the function’s area under the curve and a base that stretches over the whole interval. So…</p>
<p>avg value = (total area from a to b) / ( b -a )</p>
<p>This is helpful when they don’t give you an equation, but just graphs with easily-calculated areas (like a function consisting of a semicircle and some straight lines).</p>
<p>@ kwwboarder</p>
<p>So whenever it asks for cross sections that are circles or semi circles, the top curve-bottom curve will be the diameter?</p>
<p>Yeah, as long as it’s perpendicular to the x axis. If it’s perpendicular to the y, it’s right curve minus left curve.</p>
<p>Thanks so much… I always thought it was radius like when you do the disk/washer method -_- You just saved me some points!</p>
<p>No problem. Just remember that you only use washers/discs for revolutions and not cross sections and you should be good.</p>