# U-substitution for Area of a SemiCircle

1. Sep 5, 2012

### DahnBoson

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) lets 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.

Last edited: Sep 5, 2012
2. Sep 5, 2012

### mathman

The curve is a parabola, not near a circle.

3. Sep 5, 2012

### DahnBoson

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. Sep 6, 2012

### mathman

The integral = 4/3 (you are correct).

5. Sep 6, 2012

### DahnBoson

Thanks mathman appreciate it.