Question

Let f be a continuous function on \(\mathbb{R}\) and \( F(x) = \int_{x-2^{x+2} f(t) dt \), then F'(x) is}

• A. \( f(x-2) - f(x+2) \)
• B. \( f(x-2) \)
• C. \( f(x+2) \)
• D. \( f(x+2) - f(x-2) \)

Hint:

This is a standard application of the Leibniz rule. Remember the full formula: \( \frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t) dt = f(x, b(x))b'(x) - f(x, a(x))a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x} dt \). For this problem, \(f\) depends only on \(t\), so the integral term is zero, simplifying the rule.

Answer 1

The Correct Option is D

Step 1: Problem Overview:
The goal is to differentiate an integral with variable limits, which involves the Leibniz integral rule, a more general form of the Fundamental Theorem of Calculus.

Step 2: Core Formula:
The Leibniz rule is used to differentiate \( F(x) = \int_{a(x)}^{b(x)} f(t) dt \), and it states: \[ F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]In this problem:- \( f(t) \) is the integrand.- The upper limit is \( b(x) = x+2 \).- The lower limit is \( a(x) = x-2 \).

Step 3: Step-by-step Solution:
Calculate the derivatives of the limits:\[ b'(x) = \frac{d}{dx}(x+2) = 1 \]\[ a'(x) = \frac{d}{dx}(x-2) = 1 \]Apply the Leibniz rule:\[ F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]Substitute the functions and their derivatives:\[ F'(x) = f(x+2) \cdot (1) - f(x-2) \cdot (1) \]\[ F'(x) = f(x+2) - f(x-2) \]
Step 4: Final Result:
The derivative \( F'(x) \) is \( f(x+2) - f(x-2) \).