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November 26th, 2017, 12:21 PM   #1
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Definite integral between 2 variables

I am trying to find the area of the function bounded by 2 variables, the x axis, and the function. I have no idea where to start other than whether the function is even or odd. It is neither.

Here are the bounds:

$\displaystyle y=e^{-x}$
$\displaystyle y=0$
$\displaystyle x=a$
$\displaystyle x=b$

So how do I find this definite integral between 2 variables? The limit definition seems useless because what are you supposed to take the limit to? Infinity?
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November 26th, 2017, 01:00 PM   #2
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this is just

$A = \displaystyle \int_a^b~e^{-x}~dx = \left . e^{-x} \right|_b^a = e^{-a}-e^{-b}$

as you'd expect.
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November 27th, 2017, 01:39 AM   #3
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$$\int e^{-x}\,dx=-e^{-x}+C$$
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November 27th, 2017, 04:50 AM   #4
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The "x= a" and "x= b" are not "variables". They are constants, fixed values for the variable, x.
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November 27th, 2017, 07:13 AM   #5
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But those values are themselves unknown. Isn't that enough to consider them variables?
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November 27th, 2017, 07:16 AM   #6
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They could be variables
\begin{align*}\int_{b(t)}^{a(t)} e^{-x}\,\mathrm dx &= \int_{b(t)}^c e^{-x}\,\mathrm dx + \int_c^{a(t)} e^{-x}\,\mathrm dx \\ &= \int_c^{a(t)} e^{-x}\,\mathrm dx - \int_c^{b(t)} e^{-x}\,\mathrm dx \\ &= -e^{-a(t)}+e^{-c} + e^{-b(t)}-e^{-c} \\ &= e^{-b(t)}-e^{-a(t)}
\end{align*}
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November 27th, 2017, 07:20 AM   #7
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Quote:
Originally Posted by caters View Post
But those values are themselves unknown. Isn't that enough to consider them variables?
No, because variables change their value and constants don't.

When you meet somebody whose name you don't know, you don't consider that their name changes with time or position (or anything else). You consider that they have a single fixed name that you happen not to know. And because it is fixed you assume that they don't have to constantly apply for new passports, driving licenses and bank accounts.
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