Find this integral in terms of the given integrals

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


If [itex] \displaystyle \int_0^1 \dfrac{e^t}{t+1} dt = a[/itex] then [itex]\displaystyle \int_{b-1}^b \dfrac{e^{-t}}{t-b-1} dt [/itex] is equal to

Homework Equations



The Attempt at a Solution



I used the definite integral property in the second integral

[itex]\displaystyle \int_{b-1}^b \dfrac{e^{-2b-1-t}}{b-2-t} dt [/itex]
 

Answers and Replies

  • #2
329
34
They want you to solve the second integral in terms of the a they gave you.

Try making the substitution u = t - b + 1 in the second integral.
 
  • #3
utkarshakash
Gold Member
855
13
They want you to solve the second integral in terms of the a they gave you.

Try making the substitution u = t - b + 1 in the second integral.

Thanks. But how did it occur to you that u=t-b+1 would be a good idea?
 
  • #4
3,812
92
Thanks. But how did it occur to you that u=t-b+1 would be a good idea?

You need to change the limits to 0 and 1. So our aim is to find a substitution that would do this job.
 
  • #5
329
34
Thanks. But how did it occur to you that u=t-b+1 would be a good idea?

Well, I was trying to change your second integral so it would look more like the first. In general it is useful to get rid of the constants in the denominator. And as Pranav-Arora points out that formula is handy for getting the limits of integration right.

The problem now is that with the change u = t - b + 1 you don't have the right denominator. And if you set up u to get u + 1 in the denominator you don't have the right limits of integration.

So I'm not sure this trick is the right approach. I'll think about it some more.
 
  • #6
3,812
92
Well, I was trying to change your second integral so it would look more like the first. In general it is useful to get rid of the constants in the denominator. And as Pranav-Arora points out that formula is handy for getting the limits of integration right.

The problem now is that with the change u = t - b + 1 you don't have the right denominator. And if you set up u to get u + 1 in the denominator you don't have the right limits of integration.

So I'm not sure this trick is the right approach. I'll think about it some more.

Your substitution is OK, you need to play with the given integral.

$$ \int_0^1 \dfrac{e^t}{t+1} dt = \int_0^1 \dfrac{e^{1-t}}{2-t} dt$$

Do you see now?
 
  • #7
329
34
Your substitution is OK, you need to play with the given integral.

$$ \int_0^1 \dfrac{e^t}{t+1} dt = \int_0^1 \dfrac{e^{1-t}}{2-t} dt$$

Do you see now?

Yes, after all that I made the wrong substitution in the e##^{-t}##. Thank you for your help.
 
Last edited:

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