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MissPhiladelphia
To answer this question, first
let ∫ √(t³+1) dt = g(t) + C 
Then g'(t) = √(t³+1) 

F(x) = ∫₀ˣ √(t³+1) dt = g(t) |₀ˣ = g(x) - g(0) 

Now g(x) is some function of x, while g(0) is a constant 

F(x) = g(x) - g(0) 

Differentiate both sides: 
F'(x) = g'(x) - 0 = √(x³+1) 

So you are correct, in this case, we simply replace t with x (this is not always the case) 

F'(2) = √(2³+1) = √9 = 3 

You MUST remember that when dealing with square roots, we have: 
x² = 4 -----> x = -2 or 2 
x = √4 ----> x = 2 

That's why in the quadratic formula: x = (-b ± √(b²-4ac)) / (2a), we use a ± sign in front of square root, otherwise, if we could willy-nilly assign positive and negatives value to √(b²-4ac), then we would have no need for the ± sign. 

Also, when solving x² = 4, we usually have intermediate step 
x = ± √4, where +√4 (or simply √4) = 2, and -√4 = 2
VioletAyis

F'(2) = 3

[tex]\int\limits^x_0 {\sqrt{t^3+1}} \, dt[/tex]

Using Fundamental Theorem of Calculus:

Since x = 2, you now have:

[tex]\int\limits^2_0 {\sqrt{t^3+1} } \, dt[/tex]

so F'(x) = [tex]\sqrt{t^3+1}[/tex]

--> F'(2) = [tex]\sqrt{2^3+1}[/tex] = [tex]\sqrt{9}[/tex] = 3

so F'(2) = 3

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