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May 4th, 2017, 11:27 AM   #21
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Originally Posted by Gamegeck View Post
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.
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May 4th, 2017, 12:03 PM   #22
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Originally Posted by Joppy View Post
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 $x$.
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