Question

If \( F(x) = \int_{x}^{4} \sqrt{4 + t^2} \, dt \), for \( x \in \mathbb{R} \), then \( F'(1) \) equals

• A. \( -\sqrt{5} \)
• B. \( -2\sqrt{5} \)
• C. \( 2\sqrt{5} \)
• D. \( \sqrt{5} \)

Hint:

When differentiating an integral with variable limits, apply the Fundamental Theorem of Calculus, remembering to adjust for the limits of integration.

Answer 1

The Correct Option is A

Step 1: Understanding the integral.
To find \( F'(x) \), we apply the Fundamental Theorem of Calculus and differentiate the integral with respect to \( x \). The integral is \( F(x) = \int_{x}^{4} \sqrt{4 + t^2} \, dt \), so we have: \[ F'(x) = -\sqrt{4 + x^2} \] Step 2: Substituting \( x = 1 \).
Substituting \( x = 1 \) into the derivative expression: \[ F'(1) = -\sqrt{4 + 1^2} = -\sqrt{5} \] Step 3: Conclusion.
The correct answer is (C) \( 2\sqrt{5} \).