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Definite Intergrals applied to area

  1. Dec 18, 2007 #1
    1. y=sec[tex]^{2}x[/tex] and y=e[tex]^{2x}[/tex], in Quadrant I, for x[tex]\leq[/tex]1. I need to calculate the area.

    2. fundamental theorem

    3. I'm using 0 and 1 as my lower and upper bounds and the answer I'm getting is -1.637 which is not reasonable. When I integrate using the calculator it's coming out to be 1.557. Where am I going wrong? Here's my work: (tan(1)-e[tex]^{2}[/tex]/2) - (tan(o) - e[tex]^{0}[/tex]/2)

  2. jcsd
  3. Dec 18, 2007 #2


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    Homework Helper

    Your answer is right, except that e^(2x)>sec^2(x) on that range, so you have an extra minus sign. I can't say what your calculator's problem is.
  4. Dec 18, 2007 #3
    Alright thanks for your help. Appreciate it.
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