Question

Alright, I want to see if I understand the language of these two problems and their solutions. It asks:
If F(x) = [given integrand], find the derivative F'(x).
So is F(x) just our function, and F'(x) our antiderivative?

1) F(x) = ∫(2, x) [t^(3)+1]^(1/2)dt
Let u = [t^(3)+1]
∫(2, x) (u)^(1/2)du, all terms in u so
F'(x) = ∫(2, x) 2u^(3/2)/3
F'(x) = ∫(2, x) [2(t^(3)+1)^(3/2)]/3 dt

2) F(x) = ∫(2, x) sin(t^5)dt
FTCI right? We know the derivative of sin as an integral, so:
F'(x) = ∫(2, x) -cos(t^5)dt

We just leave the answers like that with dt? No need for C as our constant? I should not mix up ∫(2, x) as an indefinite ∫, right?

Answer 1

I am not sure I follow you but
when you change from t to u
you must also change dt to du
{ like if u = t^3 + 1, then du = 3 t^2 dt
and dt = du/3 t^2
which leaves you in a mess.

You must also change the t limits, 2 to x
to u limits
if t = 2, u = 2^3+1 = 9
and if t = x then u = x^3+1