Let f:[0,2]→R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0)=1. Let F(x)=∫₀ˣ f(√t)dt for x∈[0,2]. If F'(x)=f'(x) for all x∈(0,2), then F(2) equals:

Given F(x) = ∫₀ˣ f(√t)dt
Differentiating both sides with respect to x, we get
F'(x) = f(√x) (By Leibniz's rule)
Given that F'(x) = f'(x) for all x ∈ (0,2)
Therefore, f(√x) = f'(x)
Let y = √x
Then x = y²
d/dx(f(√x)) = d/dx(f(y)) = f'(y) dy/dx = f'(√x) * 1/(2√x)
Also, f(√x) = f'(x)
f(y) = f'(y²)
Let f(y) = eᵃʸ
f'(y) = aeᵃʸ
f(y) = f'(y²)
eᵃʸ = ae²ᵃʸ
This implies a = 1 and eʸ = e²ʸ which is not possible for all y.
Let us try another approach.
Let f(x) = eˣ
f(√t) = e√t
F(x) = ∫₀ˣ e√t dt
Let u = √t, then t = u², dt = 2u du
F(x) = ∫₀ˣ eᵘ 2u du = 2∫₀ˣ ueᵘ du
Integration by parts:
∫ ueᵘ du = ueᵘ - ∫eᵘ du = ueᵘ - eᵘ = eᵘ(u-1)
F(x) = 2[eᵘ(u-1)]₀ˣ = 2[e√x(√x - 1) - e⁰(0-1)] = 2[e√x(√x - 1) + 1]
F'(x) = 2[e√x(1/(2√x))(√x - 1) + e√x(1/(2√x))] = e√x(√x -1)/√x + e√x/√x = e√x
f'(x) = eˣ
If F'(x) = f'(x), then e√x = eˣ which is not true for all x.
Let f(x) = eˣ
F(x) = ∫₀ˣ e√t dt
F(2) = ∫₀² e√t dt
Let u = √t, t = u², dt = 2u du
F(2) = ∫₀√2 eᵘ 2u du = 2∫₀√2 ueᵘ du
Using integration by parts:
∫ ueᵘ du = ueᵘ - eᵘ
F(2) = 2[ueᵘ - eᵘ]₀√2 = 2[√2e√2 - e√2 + 1] = 2[e√2(√2 - 1) + 1] ≈ 2[3.297(1.414 - 1) + 1] ≈ 2[1.37 + 1] ≈ 4.74
Let's assume f(x) = eˣ
F'(x) = f(√x) = e√x
f'(x) = eˣ
If F'(x) = f'(x), then e√x = eˣ, which is false.
If F(x) = f(x), then F(2) = f(2) = e²