# Can you apply the fundamental theorem of calculus with the variable inside the integrand?

I was wondering if I can apply the FTOC: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(t) } dt \right) =f(x)$ to an implicit function with the variable being differentiated implicit in the function.

For example: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ xf(t) } dt \right)$ or any function in the form: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(x,t)f(t) } dt \right)$ and if so, what would we get.

• Note $\int_a^x x f(t) \mathop{dt} = x \int_a^x f(t)\mathop{dt}$ Mar 6, 2016 at 9:10
• Would this be the same if we have say: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(x,t)f(t) } dt \right)$ such as $\frac { d }{ dx } \left( \int _{ a }^{ x }{ \sqrt { 1+xt } f(t) } dt \right)$ or would we have to deal with it otherwise? Mar 6, 2016 at 9:14
• May be simply $\frac{d}{dx}(\int_a^x f(x,t)dt)$. Mar 6, 2016 at 9:19
• "wander" $\ne$ "wonder". Mar 6, 2016 at 9:27
• Sorry, English is not my first language. Edited. Mar 6, 2016 at 9:28

By definition $$\frac{d}{dx}\int_a^xf(x,t)dt=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x+\Delta x,t)dt-\int_a^x f(x,t)dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}\left(f(x,t)+\frac{\partial f}{\partial x}\Delta x\right)dt-\int_a^x f(x,t)dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x,t)dt-\int_a^x f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_x^{x+\Delta x}f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[f(x,x)\Delta x + \Delta x \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt\right]$$ $$=f(x,x)+\lim_{\Delta x\to 0} \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt$$ $$=f(x,x)+\int_a^x \frac{\partial}{\partial x}f(x,t)dt$$

In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.

• Where did $\int_x^{x+\Delta x}f(x,t)dt=f(x,x)\Delta$ come from? Jan 25, 2018 at 10:37

In mathematics any question of the form:

Can we do X?

If you have proven that you can.

Can you apply theorem X to Y?

• First, thanks for replying, Second, if I may rephrase my question, we know by the FTOC that if g is a function of x: $g(x)=\int _{ a }^{ x }{ f(t)dt }$ , then $g'(x)=f(x)$ , so my question would be if instead $h(x)=\int _{ a }^{ x }{ f(x,t)dt }$, would $h'(x)=f(x,x)$ similarly as we regard x as a constant inside the integral?. Mar 6, 2016 at 9:44
• So for example if $h(x)=\int _{ a }^{ x }{ \sqrt { 1+xt } f(t)dt }$, is it true that we say $h'(x)=\sqrt { 1+x^{ 2 } } f(x)$? Mar 6, 2016 at 9:46
• I am sorry if I irritated you in some way, and thank you I think I got it. The simplest example if $\int _{ a }^{ x }{ xf(t)dt }$, and since x is irrelevant of t we can take it out to get $x\int _{ a }^{ x }{ f(t)dt }$. If I differentiate wrt x, I get $\frac { d }{ dx } \left( x\int _{ a }^{ x }{ f(t)dt } \right) =xf(x)+\int _{ a }^{ x }{ f(t) } dt$ which is not the same as xf(x) as long as $x\neq a$. Mar 6, 2016 at 9:58
• @AspiringMat: No you didn't irritate me, but I am earnestly telling you to always check every tiny detail of the conditions required for a theorem. It is often overlooked by many students who just want the answer and not understand the logical reasoning completely. As for the simplest example, I would give $\int_0^x x\ dt = [xt]_0^x = x^2$ which when differentiate with respect to $x$ gives $2x$, not $x$. Mar 6, 2016 at 10:03
• @AspiringMat: As for your example, your last phrase is not correct. It may be that $x \ne a$ but $\int_a^x f(t)\ dt = 0$. However, you got the idea, that if that integral is not zero then it gives a desired counter-example. Mar 6, 2016 at 10:05