If $F: U \to \Bbb R$ is defined by $F(y) = \int_{\alpha(y)}^{\beta(y)}f(x,y)dx$, find $F'(y)$.

Question

Let $U$ be an open subset of $\Bbb R$, let $\alpha,\beta:U \to \Bbb R$ be differentiable functions, let $V$ be an open subset of $E^2$ that for each $y \in U$ contains the entire line segment between point $(\alpha(y),y)$ and $(\beta(y),y)$, and let $f:V \to \Bbb R$ be a continuous function such that $\partial f / \partial y$ exists and is continuous. If $F: U \to \Bbb R$ is defined by $F(y) = \int_{\alpha(y)}^{\beta(y)}f(x,y)dx$, find $F'(y)$.

I know the solution is $F'(y) = \int_{\alpha(y)}^{\beta(y)} \frac{\partial f}{\partial y}(x,y)dx+f(\beta (y),y)\beta'(y) - f(\alpha(y),y)\alpha'(y) $, but I do not know why. Could someone show me how they got here. I looked on MSE for a while but I could not find anything that really helped.

Answer 1

Go back to the definition of differentiation $$ F'(y) = \lim_{h\to 0} \frac{\int^{\beta(y+h)}_{\alpha(y+h) }f(x,y+h)dx - \int^{\beta(y)}_{\alpha(y) }f(x,y+h)dx }{h} $$ The first integral is (to first order of $h$) $$ \int^{\beta(y+h)}_{\alpha(y+h) }f(x,y+h)dx= \int^{\beta(y)+\beta'(y)h}_{\alpha(y)+\alpha'(y)h }[f(x,y)+\partial_y f(x,y)h]dx $$ $$ = \underbrace{\int^{\beta(y)}_{\alpha(y)+\alpha'(y)h }[f(x,y)+\partial_y f(x,y)h]dx}_{:=I} + \underbrace{\int^{\beta(y)+\beta'(y)h}_{\beta(y)}[f(x,y)+\partial_y f(x,y)h]dx}_{:=I'} $$ Now $$ I'=f\Big(\beta(y)+h\beta'(y),y\Big)\:\beta'(y)h + O(h^2)=f(\beta(y),y)\beta'(y)h+O(h^2) $$ And $$ I= \int^{\beta(y)}_{\alpha(y) }[f(x,y)+\partial_y f(x,y)h]dx- \underbrace{\int_{\alpha(y)}^{\alpha(y)+\alpha'(y)h }[f(x,y)+\partial_y f(x,y)h]dx}_{f(\alpha(y),y)\alpha'(y)h+O(h^2)} $$ where the second integral is evaluated similarly as before. So you find your desired answer $$F'(y) = \int_{\alpha(y)}^{\beta(y)}\partial_y f(x,y)dx + f(\beta(y),y)\beta'(y)h-f(\alpha(y),y)\alpha'(y)h$$

Answer 2

Hint

Let $\Phi(u,v)=\int_a^u f(x,v)dx$. Then

$$F(y)=\Phi(\beta(y),y)-\Phi(\alpha(y),y).$$

Rest to prove that $\frac{\partial \Phi}{\partial v}\int_a^b f(x,v)dx=\int_a^b\frac{\partial f}{\partial y}(x,v)dx$.

You have that $$\frac{\Phi(u,v+h)-\Phi(u,v)}{h}=\int_a^b\frac{f(x,v+h)-f(x,v)}{h}dx.$$

Let $t>0$ such that $[v-t,v+t]\subset U$. Let $|h|<t$. Then, by mean value theorem, there is $|c_h|<h$ s.t. $$f(x,v+h)-f(x,v)=\frac{\partial f}{\partial y}f(x,v+c_h)h.$$ Moreover, since $\frac{\partial f}{\partial y}|_{[a,b]\times [v-t,v+t]}$ is continuous, it's also uniformly continuous. Let $\varepsilon>0$. Therefore, there is a $\delta\in]0,t]$ such that for all $x\in[a,b]$, $$\left|\frac{\partial f}{\partial y}(x,v+h)-\frac{\partial f}{\partial y}(x,v)\right|<\frac{\varepsilon}{b-a}$$ if $|h|<\delta$.

Finally,

$$\left|\frac{\Phi(u,v+h)-\Phi(u,v)}{h}-\int_a^b\frac{\partial f}{\partial y}(x,v)dx\right|=\left|\int_a^b\frac{\partial f}{\partial y}(x,v+c_h)dx-\int_a^b\frac{\partial f}{\partial y}(x,v)dx \right|$$ $$\leq \int_a^b \left|\frac{\partial f}{\partial y}(x,v+c_h)-\frac{\partial f}{\partial y}(x,v)\right|dx,$$

and thus, if $|h|<\delta$ we get that $$\int_a^b \left|\frac{\partial f}{\partial y}(x,v+c_h)-\frac{\partial f}{\partial y}(x,v)\right|dx\leq\frac{\varepsilon}{b-a}\int_a^bdx=\varepsilon$$ what prove the result.