Differentiation under the integral sign

In summary, we are asked to compute the derivative of the double integral \int_c^d\int_a^t f(x,y)dx dy with respect to t. Using the fundamental theorem of calculus and differentiation under the integral sign, we get the derivative to be \int_c^d f(t,y)dy. However, we must note that F is not the antiderivative of f with respect to y, so the final answer is just F(t).
  • #1
celtics2004
3
0

Homework Statement



I am asked to compute d/dt of [tex]_{c}\int^{d}[/tex] ( [tex]_{a}\int^{t}[/tex] f(x,y)dx)dy for t [tex]\in[/tex] (a,b) for a problem involving differentiation under the integral sign.

[a,b] , [c,d] are in closed intervals in [tex]\Re[/tex]
f a continuous real valued function on [a,b] x [c,d]

Homework Equations





The Attempt at a Solution



I'm stuck because I don't really get what compute d/dt means in this context. Can someone explain to me what this problem is asking?
 
Physics news on Phys.org
  • #2
Hi celtics2004!

The double integral defines a real-valued function of t, which turns out to be differentiable. So you can compute its derivative d/dt.
To compute the derivative, you will need http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign" [Broken] (as you mentioned) and the fundamental theorem of calculus.
 
Last edited by a moderator:
  • #3
Yes, it is asking you to find the derivative of that function!
[tex]\int_a^t f(x,y)dx[/tex]
is a function of both t and y. I'm going to call that g(t,y).
[tex]\int_c^d\int_a^t f(x,y)dx dy= \int_c^d g(t,y)dt[/tex]
is a function of t only, say h(t). You are being asked to differentiate that function: dh/dt.

You can use the "fundamental theorem of Calculus", that
[tex]\frac{d}{dx}\int_a^x f(u)du= f(x)[/itex].

You can also use the fact that when you are integrating with respect to another variable, you can take the derivative inside the integral:
[tex]\frac{d}{dx}\int_a^b f(x,y)dy= \int_a^b \frac{\partial f(x,y)}{\partial x} dy[/tex]
 
  • #4
Thanks for your replies. I think I may have figured it out but I'm not very confident in my solution as this subject is still a little confusing.

Let F(t,y) = [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
Then d/dt F(t,y) = d/dt [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
Using differentiation under the integral sign I get:
d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] d/dt ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
Using the Fundamental Theorem of Calculus I get:
d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] f(t,y) dy.
Again using the Fundamental Theorem of Calculus I get:
d/dt F(t,y) = F(t,d) - F(t,c).

Does this make sense or did I screw up?

Thanks again for your help
 
  • #5
celtics2004 said:
Thanks for your replies. I think I may have figured it out but I'm not very confident in my solution as this subject is still a little confusing.

Let F(t,y) = [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.

The RHS not a function of y, since you already integrated y from d to c. It should just be F(t).

Then d/dt F(t,y) = d/dt [tex]\int^{d}_{c}[/tex] ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
Using differentiation under the integral sign I get:
d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] d/dt ( [tex]\int^{t}_{a}[/tex] f(x,y) dx ) dy.
Using the Fundamental Theorem of Calculus I get:
d/dt F(t,y) = [tex]\int^{d}_{c}[/tex] f(t,y) dy.
Again using the Fundamental Theorem of Calculus I get:
d/dt F(t,y) = F(t,d) - F(t,c).

F is not the antiderivative of f with respect to y in this case, so the last step is not correct. The final answer is what you had before that (without the y on the LHS).
 

What is "Differentiation under the integral sign"?

"Differentiation under the integral sign" is a mathematical technique used to find the derivative of a function that involves an integral. It allows us to differentiate with respect to a variable that appears both inside and outside of the integral.

Why is "Differentiation under the integral sign" useful?

This technique is useful because it can simplify complicated integrals and make them easier to solve. It also allows us to find derivatives of functions that cannot be easily differentiated using traditional methods.

What are the steps for using "Differentiation under the integral sign"?

The steps for using this technique are as follows: 1) Identify the function that needs to be differentiated. 2) Rewrite the integral so that the variable of differentiation appears both inside and outside of the integral. 3) Use the chain rule to differentiate the integral with respect to the variable. 4) Integrate the resulting function to get the final answer.

Can "Differentiation under the integral sign" be used for any integral?

No, this technique can only be used for integrals that have a variable of differentiation that appears both inside and outside of the integral. It cannot be used for integrals that do not have this property.

What are some applications of "Differentiation under the integral sign"?

"Differentiation under the integral sign" has various applications in physics, engineering, and economics. It can be used to solve problems related to motion, optimization, and probability, among others.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
515
  • Calculus and Beyond Homework Help
Replies
15
Views
745
  • Calculus and Beyond Homework Help
Replies
26
Views
821
  • Calculus and Beyond Homework Help
Replies
7
Views
133
  • Calculus and Beyond Homework Help
Replies
12
Views
936
  • Calculus and Beyond Homework Help
Replies
23
Views
862
  • Calculus and Beyond Homework Help
Replies
8
Views
720
  • Calculus and Beyond Homework Help
Replies
7
Views
642
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
741
Back
Top