Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I perform the following?

$$\frac{d}{dx} \int_0^x \int_0^x f(y,z) \;dy\; dz$$

Help/hints would be appreciated. The Leibniz rule for integration does not seem to be applicable.

share|improve this question
add comment

3 Answers

Hint: Use the following fact:

Let $\phi(\alpha)=\int_{u_1}^{u_2}f(x,\alpha)dx, ~~a\leq\alpha\leq b$, where in the functions $u_1$ and $u_2$ may depend on the parameter $\alpha$. Then $$\frac{d\phi}{d\alpha}=\int_{u_1}^{u_2}f_{\alpha}dx+f(u_2,\alpha)\frac{du_2}{d\alpha}-f(u_1,\alpha)\frac{du_1}{d\alpha}$$

Here we can consider $x$ as $\alpha$ and consider $\int_0^{x}f(y,z)dy$ as $g(x,z)$.

share|improve this answer
1  
Helpful hint! +1 –  amWhy Feb 15 '13 at 18:18
add comment

The Leibniz integral rule does work.

$$\begin{align}\frac{d}{dx} \int_{0}^{x} dy \left[\int_{0}^{x} f(y,z) dz \right] &=\int_{0}^{x} f(x,z) dz + \int_{0}^{x} dy \frac{\partial}{\partial x}\left[\int_{0}^{x} f(y,z) dz \right]\\ &= \int_{0}^{x} f(x,z) dz + \int_{0}^{x} dy \left[f(y,x) + \int_{0}^{x} \frac{\partial f(y,z)}{\partial x} dz\right]\\ &= \int_{0}^{x} f(x,z) dz + \int_{0}^{x} f(y,x) dy \end{align}$$ You only need to remember when the integration limit depends on $x$, $\frac{d}{dx}$ on the integral will pick up extra terms for the integration limits. In general:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} g(x,y) dy = g(x,b(x)) b'(x) - g(x,a(x)) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial g(x,y)}{\partial x} dy$$

share|improve this answer
1  
yes, you are right. thank you!!! –  Tomas Jorovic Feb 15 '13 at 18:30
add comment

Call your integral $I(x)$. What does $I(x)$ represent? It is the integral of function $f(y,z)$ over a square with one corner at $(0,0)$ and length of side equal to $x$. So that's $I(x)$. Now, if $x$ goes to $x+\Delta x$, what is $\Delta I$? That is the integral over the area made of two little slivers that wrap around your square from $(y=x, z=0)$, up to $(y=x, z=x)$, and back to $(y=0, z=x)$. So $\Delta I=\Delta x \times (\int_0^xf(x,z)dz + \int_0^xf(y,x)dy)$. Divide $\Delta I$ by $\Delta x$ and you have your derivative.

share|improve this answer
1  
BTW, achille hui does indeed get the same result via a correct application of the Leibniz rule. –  bob.sacamento Feb 15 '13 at 18:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.