U-substitution for Area of a SemiCircle

  • Thread starter DahnBoson
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  • #1
DahnBoson
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If anyone can help i seem to have reached a breakdown somewhere down the line or am simply lacking in some knowledge about integrals as i have not fully studied integrals, if i am trying to integrate a function from (0,2) let's say ∫[-(x-1)^2+1]dx this also equals -∫[(x-1)^2]dx+∫1dx ∫1dx=x so now i just need to integrate the other function so i use u-substitution to change x-1 into u u' should=1 after that using the following du/dx times dx=du in this equation du=dx now i have -∫[(u)^2]du which equals -1/3u^3 substitute x-1 for u an i have -1/3(x-1)^3 so the whole thing looks like this -1/3(x-1)^3+x so now i plug in 2 and 0 to get 5/3-1/3 which=4/3 which is 1.3333333333 when i expected an anwser closer to 1.5 but it might be that the curvature of the graph is not as close to the the curvature of a true circle as i originally thought.

Also on a side note i chose this equation because it seems like half a circle so i figured should get an area of about half a circle with a radius of 1 also inform me if this thinking is wrong.
 
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Answers and Replies

  • #2
mathman
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The curve is a parabola, not near a circle.
 
  • #3
DahnBoson
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i thought that it would be close enough on the interval to give me a better approximation but yeah i started thinking that the curve might be too far off as well, could you tell me if the integrals are right? I wasn't a 100% on those
 
  • #4
mathman
Science Advisor
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The integral = 4/3 (you are correct).
 
  • #5
DahnBoson
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Thanks mathman appreciate it.
 

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