Question

Let \( f(x) = \int_{1^{x} \sin^2 \left( \frac{t}{2} \right) dt \). Then the value of \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:}

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{4} \)
• D. \( 1 \)
• E. \( 0 \)

Hint:

Don't waste time actually integrating the function! Most calculus problems involving a limit of an integral are testing your knowledge of the Fundamental Theorem of Calculus.

Answer 1

The Correct Option is D

Concept: The limit \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is exactly the definition of the derivative of \( f \) at the point \( \pi \), denoted as \( f'(\pi) \). According to the Fundamental Theorem of Calculus Part 1, if \( f(x) = \int_{a}^{x} g(t) dt \), then \( f'(x) = g(x) \).

Step 1: Identify the limit as a derivative.
\[ \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} = f'(\pi) \]

Step 2: Find the general derivative \( f'(x) \).
Using the Fundamental Theorem of Calculus: \[ f'(x) = \frac{d}{dx} \int_{1}^{x} \sin^2 \left( \frac{t}{2} \right) dt = \sin^2 \left( \frac{x}{2} \right) \]

Step 3: Evaluate the derivative at \( x = \pi \).
\[ f'(\pi) = \sin^2 \left( \frac{\pi}{2} \right) \] We know that \( \sin(\pi/2) = 1 \). \[ f'(\pi) = (1)^2 = 1 \]