Solution
Proof of U'(x) = U(x)
Given that
\[
U(x) = 1 + \int_0^x U(t) \, dt
\]
Show that \( U'(x) = U(x) \)
Solution
\[
U(x) = 1 + \int_0^x U(t) \, dt
\]
Solving equation (1) using Leibniz's Rule, which states:
\[
\frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(t, x) \, dt \right) = f(h(x), x) \cdot h'(x) - f(g(x), x) \cdot g'(x)
\]
In this case:
- \( g(x) = 0 \quad \text{(Lower limit)} \)
- \( g'(x) = 0 \)
- \( h(x) = x \quad \text{(Upper limit)} \)
- \( h'(x) = \frac{d}{dx}(x) = 1 \)
- \( f(t, x) = U(t) \)
So:
\[
\frac{d}{dx} \left( \int_0^x U(t) \, dt \right) = U(x) \cdot 1 - U(0) \cdot 0 = U(x)
\]
Now differentiate equation (1):
\[
\frac{d}{dx} U(x) = \frac{d}{dx} \left( 1 + \int_0^x U(t) \, dt \right) = U(x)
\]
\[
\therefore \quad U'(x) = U(x) \quad \
