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First of all I would like to start off by asking why do they have different change of variable formulas for definite integrals than indefinite...why cant we just integrate using U substitution as we normally do in indefinite integral and then sub the original U value back and use that integrand for definite integral?

I was at one point understanding integration but not when they started coming up with different formulas for definite integrals in U-substitution I got lost and resulted to just forcibly memorizing the formulas...

I dont get why for U substitution they sub the upper and lower bounds into U from the original function to find the new upper and lower bounds with the function U.

I know that because if you dont want to sub the original value of U in and want to instead stick to U as your function you must use those new upper and lower bound but if you sub in the original value for U then you can use your old upper and lower bound values.

My question is what or how does plugging your old lower and upper bound values into U give you the new values of your new function thats expressed as U...

Why do they make such a big deal out of it and complicate it when all they have to do is same U sub as indefinite integral and then plug original value of U in and go from there...are these math people just making excuses to come up with more work or is there more logic behind it?

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    Just a note: the stated $du$ appears to be wrong, although they seem to use the correct version in the calculation.2011-12-08
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    To answer your last question, you can certainly find an indefinite integral and then evaluate at the bounds. But doing things this way can _save_ calculation, in my experience.2011-12-08
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    has anyone noticed, they forgot about the 2 from the main question when jumping to u and du2011-12-08
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    A simple answer is that if you don't change the limits, you don't have equality. If we let $u=x-2$, then whereas$$\int_0^2(x-2)\,dx \text{ is equal to }\int_{-2}^0 u\,du,$$ $$\int_0^2(x-2)\,dx\text{ is not equal to }\int_0^2u\,du.$$Also, if this were part of a long, complex computation, involving several changes of variable and/or integration by parts, are you really confident in your ability to "undo" all the substitutions and get back to the original? What if you forget? Better to keep track of those changes in the limits.2011-12-09
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    $du=-2\sin 2x\, dx\neq -\sin 2x\, dx$.2015-10-19

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