# On the Expansion of Exponential Function by Integration

1. ### lukka

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

Last edited by a moderator: Jul 19, 2013
2. ### voko

What is ##\int_0^x e^x dx ##?

3. ### lukka

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

Last edited: Jul 19, 2013
4. ### voko

Correct.

Not correct. First you find the integral, then you evaluate it.

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### Staff: Mentor

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.
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.

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6. ### voko

I agree personally, but this unfortunately tends to confuse students, in part because the ambiguous notation is well entrenched in textbooks.

7. ### lukka

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 :)

Last edited: Jul 19, 2013

### Staff: Mentor

You shouldn't do this at all if your goal is to integrate ex. See my previous post.

9. ### voko

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.

10. ### lukka

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?

### Staff: Mentor

Here: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3. There's a table of frequently used symbols about halfway down the page. Integrals are shown near the end of the table.