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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

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  1. anonymous
    • 5 years ago
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    Well you know to integrate first right? Can you do that?

  2. anonymous
    • 5 years ago
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    Yes, I already did that. The integral is arcsin(x/2).

  3. anonymous
    • 5 years ago
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    Oh 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.

  4. anonymous
    • 5 years ago
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    Yes. but since the function is undefined at the boundaries, I don't know how to set up it up when taking the limits.

  5. anonymous
    • 5 years ago
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    Undefined? Your bounds are -2 to 2

  6. anonymous
    • 5 years ago
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    This is an improper integral. I can't just plug in the boundaries.

  7. anonymous
    • 5 years ago
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    All 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.

  8. anonymous
    • 5 years ago
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    No, that's not right. Thanks for trying though..

  9. anonymous
    • 5 years ago
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    you mean the limit of the integral is from negative infinity to infinity?

  10. anonymous
    • 5 years ago
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    Because 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.

  11. anonymous
    • 5 years ago
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    I'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.

  12. anonymous
    • 5 years ago
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    The 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.

  13. anonymous
    • 5 years ago
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    branler: Thanks so much. But why is it that you took the limit approaching from the left?

  14. anonymous
    • 5 years ago
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    The 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).

  15. anonymous
    • 5 years ago
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    What about the other boundary? There is a discontinuity at x = -2 also. How do i deal with that?

  16. anonymous
    • 5 years ago
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    Do you know what the graph of 1/sqrt(4 - x^2) looks like? It's symmetrical across the y-axis. 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.

  17. anonymous
    • 5 years ago
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    Yes, 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.

  18. anonymous
    • 5 years ago
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    I 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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