Integration by part not working

  • Thread starter nysnacc
  • Start date
  • Tags
    Integration
In summary, the integrals given are from 0 to 1 and involve exponents and the variable of integration is t. The first two integrals are not integrable due to the exponent ex3-x4, but the last integral can be solved using integration by parts. The correct form of the last integral is ∫ 2xe^{x^3 - x^4}dx.
  • #1
nysnacc
184
3

Homework Statement


I have three integrals, from 0 to 1
∫ -4x5 ex3-x4dt
∫ 3x4ex3-x4dt
∫ 2tex3-x4dt

Homework Equations


Looks like they are not integrable, as ex3-x4 is not,
I tried by part, let say u =

The Attempt at a Solution


P_20161027_091741.jpg
 
Physics news on Phys.org
  • #2
nysnacc said:

Homework Statement


I have three integrals, from 0 to 1
∫ -4x5 ex3-x4dt
∫ 3x4ex3-x4dt
∫ 2tex3-x4dt

Homework Equations


Looks like they are not integrable, as ex3-x4 is not,
I tried by part, let say u =

The Attempt at a Solution


View attachment 108079
If you have written the integrals exactly as they were given, they should all be very easy. The variable of integration in all three is t (based on the dt part).

As an example, ##\int e^x dt = e^x \int 1 dt = e^x (t + C)##
 
  • #3
Oh, written wrong, should be dx,sorry
 
  • #4
Should the last one you listed be ##\int_0^1 2xe^{x^3 - x^4}dx##?

Can you post a photo of the problems? I'm wondering if there's something you left out.
 
  • #5
Yes, thanks I solved it tho, by recombine the parts in the integral and using by part
 

FAQ: Integration by part not working

What is integration by parts?

Integration by parts is a technique used in calculus to evaluate integrals of products of functions. It involves choosing one function as the "u" function and another as the "dv" function, and then using the formula ∫udv = uv - ∫vdu to find the integral.

Why is integration by parts not working?

Integration by parts may not work for certain integrals because it relies on finding a "u" and "dv" function that can be easily integrated. If these functions cannot be easily integrated, the method will not be effective.

How can I tell if integration by parts will work on a certain integral?

There are a few signs that integration by parts may be successful on a given integral. These include the presence of a product of functions, the presence of an inverse trigonometric function, or the presence of a logarithmic function.

What should I do if integration by parts is not working?

If integration by parts is not working on a given integral, there are other techniques that can be used, such as u-substitution or partial fraction decomposition. It may also be helpful to try a different choice for the "u" and "dv" functions.

Are there any common mistakes to watch out for when using integration by parts?

Yes, there are a few common mistakes that can be made when using integration by parts. These include incorrect integration of the "u" and "dv" functions, forgetting to include the negative sign in the formula, and not simplifying the resulting integral properly.

Similar threads

Replies
1
Views
1K
Replies
5
Views
992
Replies
9
Views
1K
Replies
3
Views
1K
Replies
18
Views
3K
Replies
15
Views
1K
Back
Top