Question

If F(x)=integral of sqrt(t^3+1) dt from 0 to x, then F'(2)=?

Answer 1

F'(2) = 3

`\int\limits^x_0 {√(t^3+1)} \, dt`

Using Fundamental Theorem of Calculus:

Since x = 2, you now have:

`\int\limits^2_0 {√(t^3+1) } \, dt`

so F'(x) =
`√(t^3+1)`

--> F'(2) =
`√(2^3+1)` =
`√(9)` = 3

so F'(2) = 3

Answer 2

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