Using the Fundamental Theorem of Calculus solve:
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
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
To proceed with this question, you have already computed the derivative of the integral using the Fundamental Theorem of Calculus.
Now, to find F'(x), we need to evaluate the derivative of F(x) = ∫ Cos(t^2) + t dt with respect to x.
Since F(x) is an integral with respect to t, we need to apply the chain rule.
The chain rule states that if we have an integral of the form ∫ f(g(t)) * g'(t) dt, the derivative with respect to x is given by d/dx ∫ f(g(t)) * g'(t) dt = f(g(x)) * g'(x).
In this case, our f(t) is Cos(t^2) + t and our g(t) is t. So, applying the chain rule, we have:
F'(x) = [Cos(g(x)^2) + g(x)] * g'(x)
Substituting g(x) with x, we get:
F'(x) = [Cos(x^2) + x] * 1
Thus, F'(x) = Cos(x^2) + x
Therefore, the derivative of F(x) is F'(x) = Cos(x^2) + x.
Now you have found the derivative of F(x) using the Fundamental Theorem of Calculus and the chain rule.