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Why do the lengths of the dθ segments change, while the lengths of dx do not. I can see that the x-axis is a straight line, and the half-circle is curved, but I thought the segments of dθ would be the same length -- like dividing a line or function into infinitesimally small segments for integrating.

My integral is ∫(sinθ)dθ [0, pi]. I expected the segments of dθ to be equal. Intuitively, I thought that the average height (y value) with respect to arc length would be less than the continuous average. It is. And, intuitively, I thought the average would be weighed down by the near vertical sides of the half-circle.

The only connection I have made to the reason why the length of the segments of dθ change are because the extend a horizontal length of the length of the segment of dx. But my integral is with respect to dθ. Following that reasoning, why would the segments of dθ correspond to lengths of dx.

I understand that the average with respect to arc length is smaller because of the 1/b-a factor for an average.

schematic

Bernard
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Renzo M-Svartz
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  • Why do you say the lengths of the $d\theta$ segments change? Why does it matter? The value of a Riemann integral doesn't care whether all segments are the same length, only that the lengths of all segments eventually go to zero as you make a more refined mesh. – David K Mar 15 '20 at 22:26
  • What is a "continuous" average? It seems to me that the average with respect to arc length is as "continuous" as an average ever could be, since the arc is a continuous curve, but you seem to think there is something more "continuous" than that. – David K Mar 15 '20 at 22:28
  • If you are trying to understand someone's explanation of the solution of a problem (or explanation why they think your solution was wrong), it would probably help if you included the original problem in your question text, exactly as it was given to you, and the full text of the explanation that confused you, again exactly as it was given to you. Trying to paraphrase something you don't fully understand doesn't work. – David K Mar 15 '20 at 22:35
  • @DavidK MIT single variable calculus, Clip 3, time @ 10:20 https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-60-integrals-and-averages/ – Renzo M-Svartz Mar 17 '20 at 12:36
  • I don't see anything there that says whether the $d\theta$ segments are the same or different from each other. It does say that the $d\theta$ segments are different from the $dx$ segments that you would use if you were taking an average height of the arc averaged over change in $x$ coordinate. The lecturer gives the idea that the near-vertical sides of the arc will thereby get more weight when they are averaged over $\theta$ than when averaged over $x$, which is just another way of stating your intuition that these parts of the arc would "weigh down" the average. – David K May 28 '20 at 13:12
  • If you still want an answer you might take the trouble to transcribe the particular part of the lecture that confused you, and explain exactly which statement is troublesome. This task is made easier since the site provides a transcription of the lecture: https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-60-integrals-and-averages/clip-3-average-height-with-respect-to-arc-length/R9a_NHXrBcg.pdf – David K May 28 '20 at 13:15
  • If you do copy part of the transcription, make sure it is clearly quoted in your question (not in comments--use the "edit" button under the question), and use MathJax on the formulas. – David K May 28 '20 at 13:17

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