Thread: Diff.eqn. help View Single Post
 P: 173 As I stated in my previous posts, you are not using the standar fundamental theorem of calculus, because the function you are trying to derive depends on $x$ on both the limits of integration, as on the integrand. Thats why there is an extra integral term (think of it as some sort of chain rule). The Fundamental Theorem of Calculus (roughly) states that $$\frac{d}{dx}\left(\int_a^x f(u)du\right)=f(x).$$ In your case, the integrand does not only depends on $u$, but also on $x$. Thats why there is the extra term. In a previous post, I've proven that $$\frac{d}{dx} \left(\int_a^x f(u,x)du\right)=f(x,x)+\int_a^x \frac{\partial f}{\partial x}(u,x) du.$$ In your case, lets write $y(x)$ as $$y(x)=x+\frac{1}{2}\left\{\int_{-1}^x f(u,x)du +\int_x^1 g(u,x)du\right\},$$ where $f(u,x)=[1-(x-u)]y(u)$ and $g(u,x)=[1-(u-x)]y(u)$. Using my previous posts, the derivative of $y(x)$ is $$y'(x)=1+\frac{1}{2}\left\{f(x,x)+\int_{-1}^x\frac{\partial f}{\partial x}(u,x)du-g(x,x)+\int_x^1 \frac{\partial g}{\partial x}(u,x)du\right\}.$$ Substituting $f,\,g,\,\frac{\partial f}{\partial x},\,\frac{\partial g}{\partial x}$, you should come up with the correct answer.