1. ## Differentiating an integral

Hi,

I am unsure how to differentiate an integral. The specific problem I am working on at the moment is:

$\frac{\partial }{\partial T} \int_t^T F(t;s) ds$

s is a dummy variable used to represent T. The solution to this is

$F(t;T)$

...but I am unsure why we would get this.

Peter

2. That's the first fundamental theorem of calculus: Fundamental theorem of calculus - Wikipedia, the free encyclopedia

If there's anything you should ever remember from calculus, it's this theorem.

3. Hi !
Originally Posted by peterpan
Hi,

I am unsure how to differentiate an integral. The specific problem I am working on at the moment is:

$\frac{\partial }{\partial T}\int_t^T F(t;s) ds$

s is a dummy variable used to represent T. The solution to this is

$F(t;T)$

...but I am unsure why we would get this.

Peter
Let $\tilde{F}(s)=F(t;s)$. Let $\mathcal{F}$ be an antiderivative of $\tilde{F}$.
By definition of the integral (I think you call it "fundamental theorem of calculus"), we have :

$\int_t^T \tilde{F}(s) ~ds=\mathcal{F}(T)-\mathcal{F}(t)$.

Now, $\frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\frac{\partial}{\partial T} ~ \left(\mathcal{F}(T)-\mathcal{F}(t)\right)$

Assuming that t is not a function of T, we can thus consider that $\mathcal{F}(t)$ is a constant with respect to T. Hence $\frac{\partial}{\partial T} ~\mathcal{F}(t)=0$

$\therefore \frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\frac{\partial}{\partial T} ~\mathcal{F}(T)$

But the derivative of $\mathcal{F}$ is $\tilde{F}$. We can then say that $\frac{\partial}{\partial T} \int_t^T F(t;s) ~ds=\frac{\partial}{\partial T} \int_t^T \tilde{F}(s) ~ds=\tilde{F}(T)=F(t;T)$

4. Here's the general expression:

$\frac{\partial}{\partial x}\int_{f_1(x)}^{f_2(x)}G(x,t)dt=
G(x,f_2(x))\frac{d f_2}{dx}-G(x,f_1(x))\frac{d f_1}{dx}+\int_{f_1(x)}^{f_2(x)}
\frac{\partial G}{\partial x}dt$

Then:

$\frac{\partial}{\partial x}\int_{4x+x^2}^{\sin(x)} (2x^2\sin(t)+t^2e^{x})dt$

is a piece of cake right?