Using the Fundamental Theorem of Calculus solve:

Question

F(x)= ∫ Cos(t^2) + t dt on interval [x to 2]

FTC = [∫f(x)]' on interval [0 to x]

Switching a and b change the sign of the function, since we want to x to be the upper bound, then we exchanged the t terms for the value of the upper bound which is x.

f'(x) = - [∫ Cos(t^2) + t dt on interval [2 to x]
= -1 [Cos(x^2) + x]

Just need a little help with how proceed with this question

Answer 1

Note that if F(x) = ∫f(x) dx then
f(x) = F'(x)
Now, we know that ∫[a,x] f(t) dt = F(x)-F(a)
So, d/dx ∫[a,x] f(t) dt = F'(x) - F'(a)
but since a is constant, F'(a) = 0, so we are left with just
F'(x) = f(x)

So, if F(x) = ∫[a,x] cos(t^2) dt
F'(x) = cos(x^2)

Smilarly, if F(x) = ∫[x,a] f(t) dt
F'(x) = -f(x)

Note that this can be seen as just the Chain Rule in reverse, since if u and v are functions of x,
if F(x) d/dx ∫[u,v] f(t) dt
then
F'(x) = f(v) dv/dx - f(u) du/dx