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Basic Integration Problem

  1. Jan 10, 2012 #1
    Function given:

    F(b) = ∫e-x2dx from 0 to b

    I'm asked to estimate F(1), F(2), and F(3) using a left-sum with 3 subdivisions.

    So, I guess Δx would be 1, then, so it doesn't really matter for the purposes of solving this problem. However, this is as far as I've gotten, and I haven't really been able to make much progress.

    Can anyone suggest another step I can take to come closer to the solution?

    Thank you!
  2. jcsd
  3. Jan 10, 2012 #2
    so you want to estimate the area under this curve with 3 rectangles. So what is the width of the rectangle. And how tall is the rectangle?
  4. Jan 10, 2012 #3
    The width of the rectangles would simply be 1, so they don't really matter in our calculations (i.e. we can simply add the heights together). I think the height of each rectangle would be e-x2 evaluated at 0, 1, and 2, but I'm unsure. Am I on the right track?
  5. Jan 10, 2012 #4
    yes your on the right track. the value of the function at an x would be the height.
  6. Jan 10, 2012 #5

    F(1) = e-12
    F(2) = F(1) + e-22
    F(3) = F(2) + e-32

  7. Jan 10, 2012 #6
    and should multiply each by b/3 to get the area since it's an integral.so [tex] \frac{b}{3}(e^{-1}+e^{-4}+e^{-9})=F(3)~F(b)[/tex]
  8. Jan 10, 2012 #7
    F(1) means b = 1.

    so you need to find ∫e-x2dx from 0 to 1.

    left hand estimate with 3 divisions. the width of the rectangles does matter.

    for F(1) you need to divide the space between 0 and 1 into 3 sections. I will start you off: 0, [itex]\frac{1}{3}[/itex],.....(obviously there are 2 more in order to divide into 3 equal sections).

    provided that makes sense, you can then move on to find the height of each rectangle:
    the height is going to be found by plugging in your LEFT-HAND values of x you found when you divided the graph into 3 parts.

    since I told you 0 is one of your x-values, the height for 0 will be:

    e-02 = 1

    the resulting area is going to be the (width of the rectangles) x (the sum of the heights).
    Last edited: Jan 10, 2012
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