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Math Help - [SOLVED] Limit of an Integral

  1. #1
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    [SOLVED] Limit of an Integral

    Evaluate:
     \lim_{h \to 0} \frac{1}{h} \int_0^h {5e^{-x^2}dx}

    Let  f(x) = 5e^{-x^2} , since f is continuous on [0, h] assuming h>0, I used the Fundamental Theorem of Calculus.

     F'(x) = \lim_{h\to0 } \frac{1}{h} \int_x^{x+h} f(t)dt = f(x)

    Let x = 0, we get

     \rightarrow \lim_{h\to0} \frac{1}{h} \int_0^h5e^{-x^2}dx = f(0)

     \rightarrow f(0) = 5e^0 = 5, correct?
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Evaluate:
     \lim_{h \to 0} \frac{1}{h} \int_0^h {5e^{-x^2}dx}

    Let  f(x) = 5e^{-x^2} , since f is continuous on [0, h] assuming h>0, I used the Fundamental Theorem of Calculus.

     F'(x) = \lim_{h\to0 } \frac{1}{h} \int_x^{x+h} f(t)dt = f(x)

    Let x = 0, we get

     \rightarrow \lim_{h\to0} \frac{1}{h} \int_0^h5e^{-x^2}dx = f(0)

     \rightarrow f(0) = 5e^0 = 5, correct?

    It looks dandy to me.

    Tonio
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  3. #3
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    Ah, ok, thank you. I wasn't sure if I was applying the FTC properly.
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