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Finding the exact value of an integral without calculator

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the exact value of: ∫-55 √25-x2 dx without using your calculator. (Hint: Consider the geometric significance of the definite integral.)


    2. Relevant equations
    integrate and find antiderivative and evaluate at A and B.


    3. The attempt at a solution
    -55 √25-x2 dx
    =[2/3(25-x2)3/2*2x]-55
    =(2/3(25-52)3/2*2*5)-(2/3(25--52)3/2*2*-5)
    =0-0

    I don't think I got the right answer, I thought i always find the antiderivative to integrate.
    Is there something I am doing wrong? Any help would be appreciated!
     
  2. jcsd
  3. May 11, 2012 #2

    Mentallic

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    Homework Helper

    Yes it's definitely wrong, and also it's very hard to follow what you did to tell you where you went wrong.

    What does the graph [itex]y=\sqrt{1-x^2}[/itex] represent?

    And have you learnt about using substitutions to solve integrals yet?
     
  4. May 11, 2012 #3
    thank you for your quick response,
    is [itex]y=\sqrt{1-x^2}[/itex] a circle?, oh a circle isn't continuous so i cant do antiderivatives?
    i have learned substitutions but if i let u be 25-x2, du = -2x dx, i would have to put a (-1/2x) in front of the integral, then integrate u to be (2/3)u3/2, wouldn't i still end up with (25-(52))-(25-(-5)2))
    would i just use the area of a circle formula ∏r2 for this problem?
     
  5. May 11, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A circle would be [itex]x^2+ y^2= 1[/itex]- y can be both positive and negative. [itex]y=\sqrt{1- x^2}[/itex] requires that y be non-negative- it is a semi-circle.

    Yes, that is the whole point of hint- the area under the semi-circle is half the area of a circle of radius 1.
     
  6. May 11, 2012 #5
    thank you, that makes sense now. the radius is 5 so it would be (1/2)∏52.
     
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