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November 29th, 2014, 02:34 PM   #1
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Fundamental Theorem

How do I approach these problems? I only know how to solve C.
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November 29th, 2014, 03:00 PM   #2
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if u is a function of x and a is a constant ...

$\displaystyle \frac{d}{dx} \int_a^u f(t) \, dt = f(u) \cdot \frac{du}{dx}$
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November 29th, 2014, 03:35 PM   #3
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if u is a function of x and a is a constant ...

$\displaystyle \frac{d}{dx} \int_a^u f(t) \, dt = f(u) \cdot \frac{du}{dx}$
so for #a I can describe the region as f(x) = e^(-a^2 / 2) right?

Last edited by Omnipotent; November 29th, 2014 at 04:18 PM.
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November 29th, 2014, 04:25 PM   #4
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so for #a I can describe the region as f(x) = e^(-a^2 / 2) right?
no ... the region is the signed area under the curve $\displaystyle y = e^{-x^2/2}$ and above the x-axis from x = 0 to x = a

maybe you should look at the graph of the function.
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November 29th, 2014, 06:17 PM   #5
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Quote:
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no ... the region is the signed area under the curve $\displaystyle y = e^{-x^2/2}$ and above the x-axis from x = 0 to x = a

maybe you should look at the graph of the function.
Thank you. I'm sorry to bother you again but could you tell me if this is correct?

b) N(0) = 0 because 0 is the start of the interval

c) n'(x) = e^(-x^2 / 2)

n''(x) = -4x*e^(-x^2 / 2)

d) lim as x approaches infinity = 0
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November 29th, 2014, 11:06 PM   #6
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Quote:
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Thank you. I'm sorry to bother you again but could you tell me if this is correct?

b) N(0) = 0 because 0 is the start of the interval

c) n'(x) = e^(-x^2 / 2)

n''(x) = -4x*e^(-x^2 / 2)

d) lim as x approaches infinity = 0
b) is correct.

Check your answer for $N''(x)$. Remember$$\dfrac{d}{dx}\left(e^{-x^2/2}\right)
=
e^{-x^2/2}.\dfrac{d}{dx}\left(\dfrac{-x^2}{2}\right)$$

d) is also correct.
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November 30th, 2014, 07:16 AM   #7
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You need to say what d) tells you about $N(x)$.
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