INtegration and the exponential function

In summary: Just remember that for a geometric series, the sum is given by a/(1-r), where a is the first term and r is the common ratio. So in this case, a = a/n and r = e^(a/n). Good luck!In summary, the conversation revolves around finding the integral of ex with respect to x. The person is wondering about the format of the integral and whether to use the fundamental theorem of calculus or Riemann sums. They also discuss the pattern of rectangles formed and the correct way to write the series. Ultimately, the conversation leads to the realization that it is a geometric series and the formula for finding the sum is given.
  • #1
nobahar
497
2

Homework Statement


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


Homework Equations


[tex]\int[/tex]ex dx


The Attempt at a Solution


Should I keep it in the format y=ex 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 ex and wondered if this is because you can't pursue this option, or should I keep trying?
Thanks!
 
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  • #2
Have you learned the fundamental theorem of calculus yet?
 
  • #3
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!
 
  • #4
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.
 
  • #5
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.

[tex]\frac{a}{n}[/tex] + ([tex]\frac{a}{n}[/tex] x [tex]e^\frac{a}{n}[/tex]) + ([tex]\frac{2a}{n}[/tex] x [tex]e^\frac{2a}{n}[/tex]) ... ([tex]\frac{\ (n-1)a}{n}[/tex] x [tex]e^\frac{(n-1)a}{n}[/tex]).

The curve crosses the y-axis at one, so the first area is [tex]\frac{a}{n}[/tex] x [tex]e^\frac{0a}{n}[/tex], and [tex]e^\frac{0a}{n}[/tex] = 1. So that's why there is ([tex]\frac{a}{n}[/tex] 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 [tex]e^x[/tex], then it would correspond with any point, such that when [tex]x= \frac{ka}{n}[/tex], then y would be [tex]e^\frac{ka}{n}[/tex]. The power of e matches the x-axis. Is this right so far? I get stuck from here...
Thanks guys/gals.
 
  • #6
I think your latex would be clearer if you were to put it all in a single tex tag, like this:

[tex]\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}[/tex]

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:

[tex]\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}[/tex]

Or in sigma notation:

[tex]\sum_{k=0}^{n-1} \frac{a}{n}e^{\frac{ka}{n}}[/tex]

Can you see why this is a geometric series, and how to evaluate it?
 
  • #7
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.
 
  • #8
Glad I could be of service.
 

Related to INtegration and the exponential function

1. What is integration?

Integration is a mathematical process that involves finding the area under a curve. It is essentially the reverse of differentiation, where the derivative of a function is found. Integration is used in various fields of science, including physics, engineering, and economics.

2. How is integration related to the exponential function?

The exponential function, denoted by the function y = e^x, is commonly used in integration because it has a special property where its derivative is equal to itself. This makes integration of exponential functions much simpler compared to other types of functions.

3. What is the purpose of using integration and the exponential function?

Integration and the exponential function have various applications in science, particularly in calculating quantities such as distance, velocity, and acceleration. In physics, integration is used to find the work done by a force, while in economics, it is used to calculate the growth rate of a population.

4. Can integration and the exponential function be used to solve real-world problems?

Yes, integration and the exponential function are heavily used in solving real-world problems in various fields such as physics, biology, and economics. For example, in physics, integration is used to calculate the displacement of an object, while in biology, it is used to model population growth.

5. Are there any limitations to integration and the exponential function?

One limitation of integration and the exponential function is that they can only be used on functions that have a continuous derivative. This means that there are certain types of functions that cannot be integrated using this method. In addition, integration can be a time-consuming process, especially for complex functions.

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