The question is:
Evaluate the improper integral for a>0.
The integral is:
the integral from 0 to infinity, of e^(-y/a)dy
Can anyone help me solve this? When I try I get 'a', which apparently is incorrect. Thank you!
∫e^(-y/a) dy
= -a(e^(-y/a) from 0 to ∞
= -a/e^(y/a) from 0 to ∞
as y -->∞
a/e^(y/a) ---> 0
so the integral from above
= 0 - (a/e^0) )
= 0 - a/1
= -a
Sure, I can help you with that! To evaluate the improper integral
∫[0,∞] e^(-y/a) dy
we can use a technique called integration by substitution. Let's begin!
Step 1: Make a substitution
Let's introduce a new variable u = y/a and find its derivative du/dy.
Now, du/dy = 1/a.
Step 2: Rewrite the integral
Using the substitution, replace y and dy in the integral:
∫[0,∞] e^(-y/a) dy = ∫[?,?] e^(-u) * (du/dy) dy,
where the limits of integration need to be adjusted to reflect the new variable u.
We need to find the new limits of integration for the variable u. When y = 0, u = 0/a = 0. As y goes to infinity, u goes to infinity as well. Therefore, the new limits of integration will be [0, ∞].
Step 3: Simplify and solve the new integral
Now, let's rewrite the integral using the substitution:
∫[0,∞] e^(-y/a) dy = ∫[0,∞] e^(-u) * (du/dy) dy
= ∫[0,∞] e^(-u) * (1/a) dy
= (1/a) ∫[0,∞] e^(-u) du.
Simplifying further, we have:
∫[0,∞] e^(-y/a) dy = (1/a) ∫[0,∞] e^(-u) du.
Step 4: Evaluate the integral
The integral of e^(-u) with respect to u is a well-known integral.
∫[0,∞] e^(-u) du = -e^(-u) evaluated from 0 to ∞
= -e^(-∞) - (-e^0)
= -0 - (-1)
= 1.
Finally, we can substitute the result back into our expression:
∫[0,∞] e^(-y/a) dy = (1/a) * 1
= 1/a.
Therefore, the value of the improper integral ∫[0,∞] e^(-y/a) dy is 1/a for any positive value of a.
I hope this explanation helps you understand how to solve the improper integral. Let me know if you have any further questions!