INtegration and the exponential function

[nobahar]

Homework Statement

I'm new to integration, and was attempting the integral of e^x with respect to x.

Homework Equations

$$\int$$e^x dx

The Attempt at a Solution

Should I keep it in the format y=e^x so that I use this in the calculation of the area? Or should I convert to ln y? I got stuck at a certain point using e^x and wondered if this is because you can't pursue this option, or should I keep trying?
Thanks!

 

[Citan Uzuki]

Have you learned the fundamental theorem of calculus yet?

 

[nobahar]

Thanks for the reply.
It's not asking me to do that (that is, I'm assuming you just mean use the relationship between the derivative and antiderivative), it's asking me to actually take the sum of all the areas below and the sum of all the areas above the curve and then let the number of divisions of the specified area, say 0 to A on the x-axis, tend to infinity. so that I get a better and better approximation of the area (trap the total area between an upper limit and a lower limit). If that makes sense!

 

[Citan Uzuki]

Okay, so you have to use Riemann sums. Start by setting up the Riemann sum the same way you would for any other problem, then use the fact that the sum is a geometric series to bring it into closed form.

 

[nobahar]

Thanks Citan, I'm tryig to see where I fit the geometric series in. There's a pattern, but how do I manipulate it?
Here's my reasoning for the lower boundary, I can apply the same steps to the upper one, just slightly modified. In case the following Latex is not clear, one example is: (a/n)*(e^(a/n)). The x-axis is contained within 0 and a. n is the number of divisions of the x-axis.

$$\frac{a}{n}$$ + ($$\frac{a}{n}$$ x $$e^\frac{a}{n}$$) + ($$\frac{2a}{n}$$ x $$e^\frac{2a}{n}$$) ... ($$\frac{\ (n-1)a}{n}$$ x $$e^\frac{(n-1)a}{n}$$).

The curve crosses the y-axis at one, so the first area is $$\frac{a}{n}$$ x $$e^\frac{0a}{n}$$, and $$e^\frac{0a}{n}$$ = 1. So that's why there is ($$\frac{a}{n}$$ x 1) in the above sum (although without the x1).

Apologies for the Latex! I thinks it's understandable. Basically, those are the rectangles formed when a is divided into n parts; I figured since the y-axis is $$e^x$$, then it would correspond with any point, such that when $$x= \frac{ka}{n}$$, then y would be $$e^\frac{ka}{n}$$. The power of e matches the x-axis. Is this right so far? I get stuck from here...
Thanks guys/gals.

 

[Citan Uzuki]

I think your latex would be clearer if you were to put it all in a single tex tag, like this:

$$\frac{a}{n} + \frac{a}{n}e^{\frac{a}{n}} + \frac{2a}{n}e^{\frac{2a}{n}} + \cdots + \frac{(n-1)a}{n}e^{\frac{(n-1)a}{n}$$

And this is not quite correct. You've got the right idea -- the lower sum is indeed obtained by evaluating the function at the left endpoints of each rectangle, which are at ka/n for 0≤k≤n-1. But the width of each rectangle is a constant a/n, not ka/n. So in particular, the third rectangle would have a height of e^(2a/n) and a width of a/n, and so the third term should be a/n * e^(2a/n) -- not 2a/n * e^(2a/n) as you wrote. So the correct series is:

$$\frac{a}{n} + \frac{a}{n}e^{\frac{a}{n}} + \frac{a}{n}e^{\frac{2a}{n}} + \cdots + \frac{a}{n}e^{\frac{(n-1)a}{n}$$

Or in sigma notation:

$$\sum_{k=0}^{n-1} \frac{a}{n}e^{\frac{ka}{n}}$$

Can you see why this is a geometric series, and how to evaluate it?

 

[nobahar]

Thankyou so much! Seriously, thanks for having the patience, that's really great, I think I can take it from here (and have certainly learned from this... it's such a bad mistake!). Thanks also for the advice on the Latex, I tried doing it as one but it went completely bonkers, I think I've got to work on it a bit.

 

[Citan Uzuki]

Glad I could be of service.