Differentiation under the integral sign

[celtics2004]

Homework Statement

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

[a,b] , [c,d] are in closed intervals in $$\Re$$
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?

 

[yyat]

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" (as you mentioned) and the fundamental theorem of calculus.

 

[HallsofIvy]

Yes, it is asking you to find the derivative of that function!
$$\int_a^t f(x,y)dx$$
is a function of both t and y. I'm going to call that g(t,y).
$$\int_c^d\int_a^t f(x,y)dx dy= \int_c^d g(t,y)dt$$
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].<br /> <br /> You can also use the fact that when you are integrating with respect to <b>another</b> variable, you can take the derivative inside the integral:<br /> $$\frac{d}{dx}\int_a^b f(x,y)dy= \int_a^b \frac{\partial f(x,y)}{\partial x} dy$$[/tex]

 

[celtics2004]

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

 

[yyat]

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