# On the Expansion of Exponential Function by Integration..

by lukka
Tags: expansion, exponential, function, integration
 P: 24 I ask members here kindly for their assistance. I'm having some confusion over the process of integrating inequalities, in particular for obtaining the series expansion for the exponential function by integration. The text by Backhouse and Holdsworth (Pure Mathematics 2), shows the expansion of the exponential function by integration with the assumption that (x^n)→0 as n→∞ when |x|<1; Let the variable x lie in the range of values from 0 to c, where c is any positive constant, thus 0 < x < c since e^0 = 1, ∴ 1 < (e^x) < e^c Integrating from 0 to x gives, x < (e^x - 1) < x(e^c) The problem i'm having here is the integration step, where does the negative one come from in the middle of the inequality? Thanks. Lukka
 Thanks P: 5,484 What is ##\int_0^x e^x dx ##?
 P: 24 Thanks for popping by Voko. That's the definite integral from zero to x.. i.e. the indefinite integral evaluated at x minus the indefinite integral evaluated at 0 right? So if we evaluate at x we have; ∫ e^x dx = e^x +c1, and at zero; ∫ e^0 dx = ∫ (1) dx = x +c2 Thus the definite integral of e^x from 0 to x; [e^x +c1] - [x +c2] = [e^x - x] +c1 + c2
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P: 5,484

## On the Expansion of Exponential Function by Integration..

 Quote by lukka Thanks for popping by Voko. That's the definite integral from zero to x.. i.e. the indefinite integral evaluated at x minus the indefinite integral evaluated at 0 right? So if we evaluate at x we have; ∫ e^x dx = e^x +c1
Correct.

 and at zero; ∫ e^0 dx = ∫ (1) dx = x +c2
Not correct. First you find the integral, then you evaluate it.
Mentor
P: 20,948
 Quote by voko What is ##\int_0^x e^x dx ##?
It would be better to write this as
##\int_0^x e^t dt ##
I.e., use a dummy variable that is different from the variable in the limits of integration.
 Quote by lukka Thanks for popping by Voko. That's the definite integral from zero to x.. i.e. the indefinite integral evaluated at x minus the indefinite integral evaluated at 0 right? So if we evaluate at x we have; ∫ e^x dx = e^x +c1, and at zero; ∫ e^0 dx = ∫ (1) dx = x +c2 Thus the definite integral of e^x from 0 to x; [e^x +c1] - [x +c2] = [e^x - x] +c1 + c2
No, this isn't right. An antiderivative of ex is ex. For a definite integral you can choose the antiderivative where the arbitrary constant is zero.

So,
$$\int_0^x e^t~dt = \left. e^t\right|_0^x = e^x - e^0 = e^x - 1$$

Where you went wrong was in integrating e0. You don't need to (and shouldn't) do two integrations. Once you have your antiderivative, just evaluate it at the two limits of integration.
Thanks
P: 5,484
 Quote by Mark44 It would be better to write this as ##\int_0^x e^t dt ## I.e., use a dummy variable that is different from the variable in the limits of integration.
I agree personally, but this unfortunately tends to confuse students, in part because the ambiguous notation is well entrenched in textbooks.
 P: 24 Hey Mark, thanks for sharing your insight, i guess i will have to review the topic a little and practice a bunch of examples. Much more legible with the use of dummy variable too.. with regards to the integral when i evaluate at zero, I think i understand where i'm going wrong here.. I should just write the integral of e^0 as; ∫ e^0 dx = e^0 +c = 1 +c Is that right? ie. i cant just substitute 1 in for e^0 before i integrate? Thank you both for sharing your expertise, your help is greatly appreciated :)
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P: 20,948
 Quote by lukka ah okay now i get it, i can see where i'm going wrong here.. I should write it this way; ∫ e^0 dx = e^0 +c = 1 +c
You shouldn't do this at all if your goal is to integrate ex. See my previous post.
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P: 5,484
 Quote by lukka ∫ e^0 dx = e^0 +c = 1 +c
No, this is not correct at all.

As Mark44 said, you do not compute the indefinite integral two times. You do it just once, and then you plug in the upper and lower limits.
 P: 24 Yes. i understand what you are saying. i just integrate the function e^t like an indefinite case and than plug in the upper and lower limits of integration and obtain the difference. sorry for the confusion there and thank you for pointing this out. Where can i find the tex commands for the definite integral on the edit page?
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P: 20,948
 Quote by lukka Yes. i understand what you are saying. i just integrate the function e^t like an indefinite case and than plug in the upper and lower limits of integration and obtain the difference. sorry for the confusion there and thank you for pointing this out. Where can i find the tex commands for the definite integral on the edit page?
Here: http://www.physicsforums.com/showpos...17&postcount=3. There's a table of frequently used symbols about halfway down the page. Integrals are shown near the end of the table.

The table doesn't show definite integrals, so maybe I can add some to the table. Anyway, here's what I did. I put this in HTML code tags so that it won't render in the browser, letting you see what I wrote. Note that in the table they use [ tex ] and [ itex ] tags (without the extra spaces). You can also put $$before and after your expression (same as tex and \tex) or ## before and after your expression (same as itex and \itex). $$\int_0^x e^t~dt = \left.e^t\right|_0^x = e^x - e^0$$ The above renders as$$\int_0^x e^t~dt = \left.e^t\right|_0^x = e^x - e^0$$ P: 24 Brilliant :) thanks for the link. It doesn't look too complicated i shall begin using it right away. I found some more latex under the sigma tab on the advanced page too. thanks again Mark ;) Mentor P: 20,948  Quote by lukka Brilliant :) thanks again. I am kinda new to using the html code. i can see how that translates except for the part; \left.e^t\right why do the words: left, right appear? It's actually \left.e^t\right|_0^x This is to get the vertical bar off to the right with the two limits of integration (\right|_0^x). The other end, \left. doesn't display anything, but a \right command can't appear without a \left command. This string displays like this:$$\left.e^t\right|_0^x$$ P: 24 Yes okay i understand. That's pretty cool. So the bar is inserted at the end with the underscore to indicate the limits themselves? I have bookmarked the Latex page for future reference. The sandbox is very helpful for getting started. Might take some getting used to but i'm happy to learn it, after all it helps us all better communicate and understand each other. Is Latex a universal code? i mean can i use the same tags when visiting other forums?  Mentor P: 20,948 The underscore and caret (or circumflex) are pretty standard in LaTeX for subscripts and exponents, respectively. The same notation is used in integrals and summations and elsewhere, and limits use a similar notation. $$ \int_0^1 f(x)dx$$$$ \int_0^1 f(x)dx$$$$ \sum_{n = 1}^{\infty} \frac 1 n$$$$ \sum_{n = 1}^{\infty} \frac 1 n$$$$ \lim_{x \to \infty} f(x)$$$$ \lim_{x \to \infty} f(x)$$ P: 24 I'm beginning to get the hang of this already.. Thanks very much for all your help guys! Sci Advisor HW Helper P: 4,301  Quote by Mark44 It's actually \left.e^t\right|_0^x This is to get the vertical bar off to the right with the two limits of integration (\right|_0^x). The other end, \left. doesn't display anything, but a \right command can't appear without a \left command. This string displays like this:$$\left.e^t\right|_0^x$$This is a bit of a weird case, the more common use is with brackets. E.g. compare $$( \frac{1}{x} [ \sum_{n = 1}^N n ] )$$$$( \frac{1}{x} [ \sum_{n = 1}^N n ] )$$and $$ \left( \frac{1}{x} \left[ \sum_{n = 1}^N n \right] \right) $$$$\left( \frac{1}{x} \left[ \sum_{n = 1}^N n \right] \right)$$The nice trick is the brackets don't have to match, e.g. you can use $$\left[ \sum_{n = 1}^N n \right\}$$$$\left[ \sum_{n = 1}^N n \right\}$$And if you put a dot, you will get no bracket, e.g. $$\left\{ \sum_{n = 1}^N n \right.$$$$\left\{ \sum_{n = 1}^N n \right.$$and you can combine with superscripts and subscripts $$\left. \left( \sum_{n = 1}^N n \right)^2 \right|_{x = 0}$$$$\left. \left( \sum_{n = 1}^N n \right)^2 \right|_{x = 0}

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