Question

How do you solve multiple integrals? (If possible, don't use integral calculus as explanation. It would be much appreciated if algebra or differential calculus will be used for explanation)

Answer 1

Do the inner one first, that's all I know.

Answer 2

well i will not say u the ans.. but take help with this it will help u a lot

1. Integral from 0 to 1 of e^(1/x) / x^3 dx
So i determined its discontinuous at 0. So i wrote the limit as t-->0+ of the Integral from 0 to 1 of e^(1/x) / x^3 dx.
So i used substitution, then by parts to find the anti derivative:
(1/x)e^(1/x) - e^(1/x) evaluated from 't' to 1
So i inserted the limits of integration. Then i replaced the t's with 0 to see if it converges or not, and i have a quick question. When i evaluate the limit at 0 i get:
-(1/0)e^(1/0) + e^(1/0)
My problem is what does this simplify to?
0 + 1 therefore it is convergent? Or because we are dividing by 0's the limit does not exist? this is what im confused on here

2. Sketch the region and finds its area (if the aea is infinite: S = {(x,y) | x > 0, 0 < y < xe^(-x)}
For this one, am i basically drawing the graph of xe^(-x) in quadrant one, and evaluating the limit from 0 to infinity? I just wanted to check before i did all the work, and it turned out to be not what im supposed to do! Cause we did NO examples like this in class :/

3. Use the Comparison Theorem to determine whether the integral is convergent or divergent: integral from 1 to infinity of [2 + e^(-x)]/ x dx
To be honest i have NO idea how to use the Comparison Thereom, he went through this very fast the end of friday's class, and i'm having trouble finding good examples online. Any explanation of the comparison thereom would be greatly appreciated.
Jet1045

Answer 3

@rohan do you know any other way? like algebra or differential calculus? i dont know much about integral calculus

Answer 4

@lgbasallote its simple go through my ans.. attachment file as well!!

Answer 5

using limits of Riemanian sums.

Answer 6

@rohan unfortunately i really dont understand it =)) but i do appreciate the thought...@ishaan is it somehow like partial derivatives? @myko what?? haha lol

Answer 7

lgbasallote is alright a man like u forgot every thing!! come on i know that u can do it!!

Answer 8

the definition it self of a defind integral is a limit of Rieman summ of f(x)deltax where deltax --->0 .

Answer 9

just do it twice

Answer 10

or 3 times or what ever

Answer 11

@_@ wowww i dunno what you guys are saying :P hahahaha honestly...all i know about integral calculus is substitution, trigonometric substitution, integration by parts so in short...the basics =)) is there really no way to explain this in algebra or differential calculus? coz those are the things that i have mastered..

Answer 12

diferential calculus is about limits.
\[\int\limits_{a}^{b}f(x) = \lim_{\Delta x \rightarrow 0} \sum_{n=1}^{m}f(x)\Delta x\]
so as you can see only limits and some sums are involved. But that a preaty tough way to deal with the integral by its definition

Answer 13

i could see that it is hard =))) anyway ill try learning the other aspects of integral calculus first..maybe that would be a better idea =))))