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anonymous
 5 years ago
How do I evaluate the following improper integral?
2
∫ dx / sqrt(4  x^2).
2
I know that the integral is arcsin(x/2) and that the area beneath the curve and between the boundaries is equal to pi. I also know that the function is undefined at the boundaries and they should each be replaced with a constant and that the limit should be taken when after plugging in the limits. However, I'm getting stuck. Can someone please show me the steps after finding the antiderivative? Thanks
anonymous
 5 years ago
How do I evaluate the following improper integral? 2 ∫ dx / sqrt(4  x^2). 2 I know that the integral is arcsin(x/2) and that the area beneath the curve and between the boundaries is equal to pi. I also know that the function is undefined at the boundaries and they should each be replaced with a constant and that the limit should be taken when after plugging in the limits. However, I'm getting stuck. Can someone please show me the steps after finding the antiderivative? Thanks

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well you know to integrate first right? Can you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I already did that. The integral is arcsin(x/2).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh woops :) Ok so now you want to find out the area under the curve between the x axis and curve right? Thats what integrating is, finding the area of a boundry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. but since the function is undefined at the boundaries, I don't know how to set up it up when taking the limits.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Undefined? Your bounds are 2 to 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is an improper integral. I can't just plug in the boundaries.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All you should have to do is put in arcsin(2/2)arcsin(2/1)... am I off here? I'm not a pro at Calc myself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, that's not right. Thanks for trying though..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean the limit of the integral is from negative infinity to infinity?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because the area under the "left" side is equal to the area on the "right" side, just worry about the right side (we can multiply that area by 2 when we find it). Evaluate the integral from 0 to b, you should get arcsin(b/2)arcsin(0). Then, find...\[\lim_{b \rightarrow 2^}\arcsin(b/2)\] This value is pi/2. Then multiply this area by 2 to get pi for the whole area.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm learning from the book to replace the boundaries something like this: ................u arcsin(x/2)[ = arcsin(u/2)  arcsin(1) = arcsin(u/2)  3pi/2 ................2 Then to take the limit.. lim u>2 [ arcsin(u/2)  3pi/2 ] .... .. and do the same for the other boundary. I know the answer is pi, but I don't know how to arrive at that answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The way I look at it is that you have a "problem" at x = 2 because there is a discontinuity. To change it, replace the 2 with a b, u, or whatever letter and find the limit as b approaches that number of the whole integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0branler: Thanks so much. But why is it that you took the limit approaching from the left?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limit approaches from the left because values greater than 2 (or to the right of 2) are not in the domain of arcsin(x/2).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What about the other boundary? There is a discontinuity at x = 2 also. How do i deal with that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know what the graph of 1/sqrt(4  x^2) looks like? It's symmetrical across the yaxis. Therefore, you can find the area of one side (I chose the right side), then multiply that by 2 to get the whole area. If you want to know how to do it, though, replace the 2 with an a, b, u, or whatever letter (but not the same as the other one) and find the limit of the entire integral as that value a approaches 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I know what it looks like. But I much rather solve it analytically than by looking at a visual representation. Thanks so much for your help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't why you are saying that the integral is not defined in the borders.. arcsin(1), and arcsin(1) are both defined and equal to pi/2 and pi/2, respectively. that's your question can be solved with direct substitution as arcsin(1)arcsin(1)=pi/2(pi/2)=pi. Note: arcsin(x) is defined at 1<=x<=1 isn't that right?
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