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May 4th, 2017, 11:27 AM   #21
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 Originally Posted by Gamegeck I think of the indefinite integral as a general formula for the area between two bounds. Lets look at f(x)=2x for simplicity. Its defined over all real numbers. The integral(anti-derivative) is F(x)=x^2 You could have the integral be definite with some bounds in f's domain, say 3 to 5 and get the area under the curve between 3 and 5 right away. Or you could take the indefinite integral and get the anti-derivative. From there you can evaluate F(5)-F(3) and get the same area. So as you can see its a more general form of the area. Hence your intuition about dx being the width of a very thin rectangle can be applied to indefinite integrals. Its just more of a formula now.
You should beware of regarding an integral as an area; in appropriate circumstance it indeed equates to an area but in others taking the integral will yield the wrong answer. May 4th, 2017, 12:03 PM   #22
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 Originally Posted by Joppy What do we do in this case.
Pick a better $u(x)$ or be very careful when converting back from $u$ to $x$. The only place I've seen it matter is in trigonometric substitutions where there is always another option.

The problem arises because in going from $u$ to $x$ we need the inverse function $u^{-1}$, but this is not well defined if $u$ is not monotonic. You may have to make sure that you are picking the correct value for $x$. Tags calculus, indefinite, indefinite integrals, integrals 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 Sefrez Calculus 1 March 8th, 2013 10:19 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM The Chaz Calculus 1 August 5th, 2011 09:03 PM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

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