Exact form $\alpha$ and $dg=\alpha$

Let $$\alpha= f_1(x_1,x_2)dx_1 + f_2(x_1,x_2)dx_2$$ such that $$d\alpha=0$$. Define a function $$g$$ by $$g(x_1,x_2)= \int_{0}^{x_1} f_1(t,x_2) dt + \int_{0}^{x_2} f_2(0,t)dt$$ Show that $$dg=\alpha$$

By $$d\alpha=0$$ we get $$\dfrac{\partial{f_2}}{\partial{x_1}}-\dfrac{\partial{f_1}}{\partial{x_2}}=0$$ and $$dg= \dfrac{\partial}{\partial{x_1}} ( \int_{0}^{x_1} f_1(t,x_2) dt ) dx_1+\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_1} f_1(t,x_2) dt) dx_2$$

$$+\dfrac{\partial}{\partial{x_1}}(\int_{0}^{x_2} f_2(0,t) dt) dx_1 +\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_2} f_2(0,t) dt) dx_2$$ By Leibniz integral rule

$$\dfrac{\partial}{\partial{x_1}} ( \int_{0}^{x_1} f_1(t,x_2) dt ) =f_1(x_1,x_2) + ( \int_{0}^{x_1}\dfrac{\partial}{\partial{x_1}} f_1(t,x_2) dt )$$

$$\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_1} f_1(t,x_2) dt)=(\int_{0}^{x_1} \dfrac{\partial}{\partial{x_2}}f_1(t,x_2) dt)$$

$$\dfrac{\partial}{\partial{x_1}}(\int_{0}^{x_2} f_2(0,t) dt)=(\int_{0}^{x_2} \dfrac{\partial}{\partial{x_1}}f_2(0,t) dt)$$

$$\dfrac{\partial}{\partial{x_2}}(\int_{0}^{x_2} f_2(0,t) dt)=f_2(0,x_2)+(\int_{0}^{x_2}\dfrac{\partial}{\partial{x_2}} f_2(0,t) dt)$$

Do i have any mistake? And can you continue to compute those and indicate $$dg=\alpha$$?

• You only need to use Leibniz's integral rule 1 time and not 4 times. Check out my comment to this question. It's all about understanding what a single integral of a multivariable function is. Once you see this, then use what we know from single variable calculus: $\frac{d}{dx}\int_a^x f = f(x)$. Therefore your first line, 3rd line, and 4th line under the link you give is just applying this result. Only the 2nd line do you need leibniz Commented Oct 3, 2018 at 19:25

This is fine so far. Now $$\frac{\partial}{\partial x_1} f_1(t,x_2) = \frac{\partial}{\partial x_1} f_2(0,t) = \frac{\partial}{\partial x_2} f_2(0,t) = 0.$$ Why? All that you have left to do is use the hypothesis that $$d\alpha=0$$ to rewrite $$\int_0^{x_1} \frac{\partial}{\partial x_2} f_1(t,x_2)dt = \int_0^{x_1}\frac{\partial}{\partial x_1} f_2(t,x_2)dt,$$ and finish the computation.