Solving Fundamental theorem of calculus sums by differentiating the integral directly.

Question

We know the Fundamental theorem of calculus part 1 states that,

F'(x) = d/dx INTEGRAL from a to x f(t) dt = f(x)

when we solve sums

For eg d/dx [integral from 0 to √x cos t dt] we get

= (cos √x) (d/dx (√x))

= cos√x /2√x

Why do we put the value OF the upper bound of the integral in place of t, when we differentiate the function??

I know the proof of FTC and F'(x)=f(x), i know what it is.

i dont know if this is a silly question.forgive me

thank you.

Answer 1

I am integrating t=a to t=x of f(t)dt. Let u=f(t), du/dt=f'(t), when t=a, u=f(a), when t=x u=f(x). dt=du/f'(t)

so the integral is same as integral u=f(a) to u=f(x) of du/f'(t) where f'(t) is a function of u (need to make substitution).

ex. integral t=0 to t=x of t²dt same as integral from u=0 to u=x² of (1/2)u^(1/2) where u=t² ,,,,t=u^(1/2) du/dt=2t

so dt=du/2t and like mentioned t=u^(1/2),,,

Does that help?