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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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    @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.2012-12-06