Question

If \( f(x) = \int_0^{x^3} (t + 4)^2 \, dt \), then \( f'(2) \) is equal to:

• A. 288
• B. 432
• C. 144
• D. 216
• E. 24

Hint:

Use Leibniz's rule to differentiate integrals with variable upper limits.

Answer 1

The Correct Option is B

We are given \( f(x) = \int_0^{x^3} (t + 4)^2 dt \). 
To find \( f'(x) \), we apply the Leibniz rule for differentiation under the integral sign: \[ f'(x) = \frac{d}{dx} \left( \int_0^{x^3} (t + 4)^2 dt \right) = (x^3)' \cdot (t + 4)^2 \Big|_{t = x^3}. \] Thus, we have: \[ f'(x) = 3x^2 \cdot (x^3 + 4)^2. \] Now, evaluating at \( x = 2 \): \[ f'(2) = 3(2)^2 \cdot (2^3 + 4)^2 = 3 \cdot 4 \cdot (8 + 4)^2 = 3 \cdot 4 \cdot 12^2 = 432. \] Thus, \( f'(2) = 432 \).