#FTC
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When you have a change of variables (such as here, t -> x) you need to be careful how you deal with it. If the upper limit of the integral was just x you could get away with direct substitution
Let me give you an easy example. If you had $\frac{d}{dx}\int_0^{2x} t^3\text{d}t$ the answer would not be $\frac{d(8x^3)}{dx}=24x^2$ but rather $16x^3$
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Tiago
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Try applying a change of variables such as $2x=u(x)$ and going through the math, remembering the rules of derivatives throughout. So you would start with $\frac{d}{dx}\int_0^{u} t^3\text{d}t=u^3\frac{du}{dx}$
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Tiago
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Yours is more complicated but hopefully if you understand this one you'll understand why your initial statement was wrong
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Try applying a change of variables such as $2x=u(x)$ and going through the math,...
If I understood you correctly and after some long thinking, according to the FTC, $\frac{d}{dx} \int_{a}^{x}{f(t)dt} = f(x)$ can be reasoned as the derivative of $\frac{d}{dx}(\left F(x)-F( a)\right) $. The derivative of $-F(a) = 0$ since it'll just be a constant term. For any top-bound of the integral, e.g $u(x)$ like you said should result in $\frac{d}{dx}(\left F(u(x))-F( a)\right) = F'(\left u(x)\right)u'(x)$
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frigid bronze
If I understood you correctly and after some long thinking, according to the FTC...
That's exactly it. So, using that logic, if you had two functions $u(x)$ and $v(x)$ and then $\int_{v(x)}^{u(x)}f(t)dt$ you would get $F'(u)u'-F'(v)v'$
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Tiago
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That's exactly it. So, using that logic, if you had two functions $u(x)$ and $v(...
Yeah, the lower bound can also be some function! That's pretty neat!!
I missed my lecture on FTC so I was struggling as we really don't have a textbook
I feel relieved now
Thank you so much Tiago for taking your time to explain and lead me to a wonderful understanding ❤️
frigid bronze
My pleasure!