Integrating x^ne^xn: Analyzing & Solving

  • I
  • Thread starter Mayhem
  • Start date
  • Tags
    Integrating
In summary, the conversation discusses the integration of the given expression and suggests using integration by parts as the best method. It also points out a mistake in the original solution, where the -1 term was incorrect due to a redundant bound in the sum. After correcting this mistake, the final evaluation is given.
  • #1
Mayhem
331
219
I was bored and tried to integrate ## x^n e^{xn} ##. I seem to be on the right track, but ultimately it is not entirely correct. Here is my work:

Given is the integral
$$I = \int x^ne^{nx}dx$$

where ##n \geq 1##

We substitute ##t = nx## which gives us ## \frac{dt}{dx} = n \Rightarrow dx = \frac{1}{n}dt## and ##x = \frac{t}{n}##

Plugging that in, we obtain
$$ I = \int \left(\frac{t}{n}\right)^ne^t \frac{1}{n}dt $$
which can be reduced to
$$ I = \frac{1}{n^{n+1}} \int t^ne^t dt $$
The expression inside of the integral can be integrated by parts if we simply integrate ##n## times, letting ##dv = e^t## every single time, the degree of the t-polynomial will decrease to zero and cancel out to 1.

We may do this for the first few steps to obtain a pattern:
$$I =\frac{1}{n^{n+1}} \left( e^tt^n - \int nt^{n-1}e^t dt \right ) $$
$$=\frac{1}{n^{n+1}} \left( e^tt^n - \left (nt^{n-1}e^t - \int n(n-1)t^{n-2}e^t dt \right ) \right) $$
$$=\frac{1}{n^{n+1}} \left( e^tt^n - \left (nt^{n-1}e^t - \left ( n(n-1)t^{n-2}e^t - \int n(n-1)(n-2)t^{n-3}e^t dt \right) \right) \right)$$
Recalling that ## \int udv = uv - \int vdu##, we may create a general expression for ##n## terms. Generating the sum, it should be fairly obvious that
$$I = \frac{1}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}t^{n-m}e^t - \int e^t dt \right)$$
Imagine that we have ##n!t^0 e^t## inside of the integral. We may factor out ##n!## and ##t^0 = 1##.
We evaluate the integral, and then factor out ##e^t##
$$ I = \frac{e^t}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}t^{n-m} - 1 \right) + C$$
And we substitute ##t=nx## back in
$$ I = \frac{e^{nx}}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}(nx)^{n-m} - 1 \right) + C$$
Where did I go wrong? It seems to particularly be the ##\frac{1}{n^{n+1}}## factor that messes up results. I tried manually evaluating for n = 1, 2, 3, 4 and compared to calculators.
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
This may help its a bit more general replacing nx with ax but should still give the same answer:



Be careful, if you make a mistake again the instructor is likely to bench press you. :-)
 
  • Like
Likes Mayhem and Adesh
  • #3
Use partial integration: [itex] \int (x^{n} e^{nx})dx=x^{n}\frac{1}{n} e^{nx}-\frac{1}{n} \int (nx^{n-1}e^{nx})dx[/itex]. Sooner or later you will end up in [itex]\int (x^{0} e^{nx})dx[/itex].

Or you can start at [itex] \int (x e^{nx})dx[/itex] and work your way up using induction.
 
  • Like
Likes Adesh
  • #4
Notice that you got a product of two functions, ##x^n## and ##e^nx##. So, from elementary classes we can say that integration by parts will be the best method here.
 
  • #5
Svein said:
Use partial integration: [itex] \int (x^{n} e^{nx})dx=x^{n}\frac{1}{n} e^{nx}-\frac{1}{n} \int (nx^{n-1}e^{nx})dx[/itex]. Sooner or later you will end up in [itex]\int (x^{0} e^{nx})dx[/itex].

Or you can start at [itex] \int (x e^{nx})dx[/itex] and work your way up using induction.
So, instead of doing a substitution, I can should do IBP from the get-go?
 
  • #6
So it seems I was ONLY wrong by the -1 term, and after having slept on it, I see my mistake. It would only be true if the bound of the sum was n-1, but that would be redundant as letting the bound equal n gives us

$$ I = \frac{e^{nx}}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}(nx)^{n-m}\right) + C$$

which concludes the evaluation. :)
 
  • Like
Likes jim mcnamara, dRic2 and jedishrfu

Related to Integrating x^ne^xn: Analyzing & Solving

What is the purpose of integrating x^ne^xn?

The purpose of integrating x^ne^xn is to find the antiderivative of the given function. This allows us to solve for the original function and evaluate it at specific values.

What are the steps for integrating x^ne^xn?

The steps for integrating x^ne^xn are as follows: first, use the power rule to rewrite the function as x^n times e^xn. Then, use integration by parts to solve for the antiderivative. Finally, evaluate the antiderivative at specific values to find the solution.

What are the common mistakes made when integrating x^ne^xn?

Some common mistakes when integrating x^ne^xn include forgetting to apply the power rule, making errors in the integration by parts process, and forgetting to evaluate the antiderivative at specific values. It is important to double check each step and be careful with algebraic manipulations.

What are some real-world applications of integrating x^ne^xn?

Integrating x^ne^xn can be used in various fields such as physics, engineering, and economics. For example, in physics, it can be used to calculate the work done by a variable force, and in economics, it can be used to model the growth of a population over time.

How can integrating x^ne^xn be used to solve problems?

Integrating x^ne^xn can be used to solve problems by providing a general solution to the given function. This allows us to find the exact value of the function at specific points, which can be useful in a variety of real-world scenarios. Additionally, by understanding the process of integration, we can use it to solve more complex problems involving other functions.

Similar threads

Replies
4
Views
1K
Replies
2
Views
2K
Replies
0
Views
444
Replies
2
Views
1K
Replies
1
Views
2K
Replies
1
Views
445
Replies
16
Views
3K
Replies
5
Views
1K
  • Calculus
Replies
1
Views
1K
Replies
4
Views
1K
Back
Top