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Given that $$G(x_1,x_2) = \int_{0}^{x_1}g_1(x,0)dx + \int_0^{x_2}g_2(x_1,y)dy$$ and $$\frac{\partial g_1}{\partial x_2} = \frac{\partial g_2}{\partial x_1}$$

where $g_1 : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, $g_2 : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ How can I show that $$\frac{\partial G}{\partial x_1} = g_1(x_1,x_2)$$

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Derivation is a linear operation, so $$ \frac{\partial G}{\partial x_1} = \frac{\partial}{\partial x_1}\int_0^{x_1}g_1(x,0)\,\mathrm dx + \frac{\partial}{\partial x_1}\int_0^{x_2}g_2(x_1,y)\,\mathrm dy $$ The first term is the derivative of the integral (with respect to $x_1$), so you should know how to compute it. For the second term, you need to justify this: $$ \frac{\partial}{\partial x_1}\int_0^{x_2}g_2(x_1,y)\,\mathrm dy = \int_0^{x_2}\frac{\partial g_2}{\partial x_1}(x_1,y)\,\mathrm dy. $$ There should be a theorem somewhere in your notes/books about that. Finally, just use the property $\partial g_2/\partial x_1=\partial g_1/\partial x_2$, and then the fundamental theorem of calculus should seal the deal.

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I got to the same spot as you, and I am unsure how to proceed from there. I get that the first term is $g_1(x_1,0)$, but how to apply $\partial g_2/\partial x_1=\partial g_1/\partial x_2$ to second term? And will it really get down to $g_1(x_1,x_2)$? Given the first term it doesn't seems so. –  Sunny88 Dec 6 '12 at 6:43
    
You apply $\partial g_2/\partial x_1=\partial g_1/\partial x_2$ to the right-hand side of the second equation I wrote. Then what you get is the integral of a derivative, namely the integral w.r.t the second argument, of the derivative of $g_1$ w.r.t that same second argument. So it has the form $\int f'(y)\,\mathrm dy$ for some function $f$, and you can apply the fundamental theorem of calculus. –  jathd Dec 6 '12 at 7:34
    
@Sunny88 Also, you should mention what you have already tried and/or done in your questions. People are usually more motivated to answer when you show some work (even if it doesn't lead anywhere), and the answers are better targeted at your issue. –  jathd Dec 6 '12 at 7:36
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