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 November 26th, 2017, 12:21 PM #1 Senior Member   Joined: Nov 2013 Posts: 247 Thanks: 2 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? November 26th, 2017, 01:00 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,646 Thanks: 1476 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. November 27th, 2017, 01:39 AM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,979 Thanks: 1161 Math Focus: Elementary mathematics and beyond $$\int e^{-x}\,dx=-e^{-x}+C$$ November 27th, 2017, 04:50 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The "x= a" and "x= b" are not "variables". They are constants, fixed values for the variable, x. November 27th, 2017, 07:13 AM #5 Senior Member   Joined: Nov 2013 Posts: 247 Thanks: 2 But those values are themselves unknown. Isn't that enough to consider them variables? November 27th, 2017, 07:16 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 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*} November 27th, 2017, 07:20 AM   #7
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Quote:
 Originally Posted by caters 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. Tags definite, integral, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Johnny223 Calculus 6 February 17th, 2016 10:27 AM panky Calculus 1 April 24th, 2014 11:28 PM Random Variable Calculus 0 January 22nd, 2013 11:36 PM panky Calculus 3 January 8th, 2012 09:04 AM panky Calculus 2 August 25th, 2011 05:26 AM

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