Integration Watch

stanners 27
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#1
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can anybody help find the definate integral of f(x)=e^(-nx) from 0 to n?
thanks guys
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Mr M
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#2
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(Original post by stanners 27)
can anybody help find the definate integral of f(x)=e^(-nx) from 0 to n?
thanks guys
\displaystyle \int e^{kx} \, dx = \frac{e^{kx}}{k} + c
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stanners 27
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(Original post by Mr M)
\displaystyle \int e^{kx} \, dx = \frac{e^{kx}}{k} + c
So that gives (-e^-n2)/n between 0 and n. but then you have to divide by 0.
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Kvothe the Arcane
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(Original post by stanners 27)
So that gives (-e^-n2)/n between 0 and n. but then you have to divide by 0.
What? I think you misunderstood.
\displaystyle \int_0^n e^{-nx}dx \neq \left[\dfrac{-e^{-n^2}}{n} \right]_0^n
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Mr M
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(Original post by stanners 27)
So that gives (-e^-n2)/n between 0 and n. but then you have to divide by 0.
x is zero not n.
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stanners 27
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#6
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(Original post by Mr M)
x is zero not n.
you have to divide by x, as that is in the indefinateintegral. then replace x with the boundries. Don't you?
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stanners 27
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#7
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(Original post by Mr M)
x is zero not n.
ahh i see now. i was confused by x and n. thanks.
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Mr M
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(Original post by stanners 27)
you have to divide by x, as that is in the indefinateintegral. then replace x with the boundries. Don't you?
No you don't divide by x at any point. Look at my first post carefully. And it's "definite".
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Mr M
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(Original post by stanners 27)
ahh i see now. i was confused by x and n. thanks.
It's a fairly common mistake to make if that makes you feel any better!
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stanners 27
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#10
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(Original post by Mr M)
It's a fairly common mistake to make if that makes you feel any better!
Thanks. I don't suppose you know how to calculate infimum of f(x)=(x^0.5)/
(2+x) do you?
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Mr M
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(Original post by stanners 27)
Thanks. I don't suppose you know how to calculate infimum of f(x)=(x^0.5)/
(2+x) do you?
You could sketch the graph.
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