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Area problem

  1. Jun 26, 2005 #1
    This one has me stumped, I dont even know where to start.
    -Find the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3).
    Thanks for any help
  2. jcsd
  3. Jun 26, 2005 #2


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    Try plotting the curves, and check definite integrals.
  4. Jun 26, 2005 #3
    How would I go about checking for definate integrals? The book we use is not very descriptive on this topic. Thanks.
  5. Jun 26, 2005 #4
    Plot the area, then find your integration limits. That's what he means by definite integral. Your final result will be a number, not an expression, representing the area in the shape.
  6. Jun 27, 2005 #5


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    If you are doing a problem like this, then you certainly should be familiar with area given by integrals- that's the whole point of a problem like this. If you graph this, you will see that the two curves intersect at x= 0 and are bounded on the right by the vertical line x= 3. If you divide the area into thin rectangles, the height of each rectangle will be e3x- ex and the width will be "dx". Now, what integral is that?
  7. Jun 27, 2005 #6


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    have you tried estimating it? i.e. can you show the area is less than 1,000?

    or 10?
  8. Jun 28, 2005 #7


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    Since this has been around for a couple of days now:

    The graph of y= e3x is always above y= ex for x> 0. The two graphs cross at x= 0 and we are told that the are is bounded on the right by x= ln 3.

    The area is [tex]\int_0^{ln 3}(e^{3x}- e^x)dx[/tex].

    Can you do that integral?
  9. Jun 28, 2005 #8
    first draw the graph of all three fuctions. you'll get the picture. find the point of intersection of e^x, e^2x and x = ln3. this can be done easily...

    one point is (0,1)...trivial solution
    put x=ln3 in e^x and get y=3........(ln3,3)
    put x=ln3 in e^2x and get y=9........(ln3,9)

    now, for the next part you have to draw the graph.

    integrate e^x limits 0 to 3 to get e^3 - 1
    integrate e^2x limits 0 to 9 to get (e^9 - 1)/2

    The area required is [(e^9 - 1)/2 - (e^3 - 1)]
    note: area under graph of a function between a and b is its definite integral with a and b as lower and upper limits.
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