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Math Help - Area A(x) rectangle under curve

  1. #1
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    Area A(x) rectangle under curve

    Let A(x) be the area of the rectangle inscribed under the curve y= e^{-2x^{2}} with vertices at (-x,0) and (x,0), x\geq 0.

    a) Find A(1).
    b) What is the greatest value of A(x)? Justify your answer.
    c) What is the average value of A(x) on the interval 0 \leq x \leq 2?

    I obtained:
    a) A(x)=2x{e}^{-2{x}^{2}};A(1)= 2{e}^{-2}
    b) A'(x)=(-4{x}^2 +1)2{e}^{-2{x}^{2}}; Since A increases on -\frac{1}{2} < x < \frac{1}{2} and decreases on \frac{1}{2} < x < \infty, there is a local maximum at x=\frac{1}{2}. A(\frac{1}{2}) ={e}^{-\frac{1}{2}}

    c)avg value = \frac{1}{2} \int_0^2 {A(x)} dx = \frac{1}{4}(1-e^{-8})

    I am uncertain if I have found the correct function for A(x).
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  2. #2
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    Quote Originally Posted by xxsteelxx View Post
    Let A(x) be the area of the rectangle inscribed under the curve y= e^{-2x^{2}} with vertices at (-x,0) and (x,0), x\geq 0.

    a) Find A(1).
    b) What is the greatest value of A(x)? Justify your answer.
    c) What is the average value of A(x) on the interval 0 \leq x \leq 2?

    I obtained:
    a) A(x)=2x{e}^{-2{x}^{2}};A(1)= 2{e}^{-2}
    b) A'(x)=(-4{x}^2 +1)2{e}^{-2{x}^{2}}; Since A increases on -\frac{1}{2} < x < \frac{1}{2} and decreases on \frac{1}{2} < x < \infty, there is a local maximum at x=\frac{1}{2}. A(\frac{1}{2}) ={e}^{-\frac{1}{2}}

    c)avg value = \frac{1}{2} \int_0^2 {A(x)} dx = \frac{1}{4}(1-e^{-8})

    I am uncertain if I have found the correct function for A(x).
    your function for A(x) is correct.
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  3. #3
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    Thanks, I do not have the solution to this problem and just wanted to make sure.
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