1. **Problem:** Find the derivative of the function \( g(x) = \int_2^x x^2 \sin x \, dx \). 2. **Formula and rule:** By the Fundamental Theorem of Calculus, if \( F(x) = \int_a^x f(t) \, dt \), then \( F'(x) = f(x) \). 3. **Important note:** Here, the integrand depends on \( x \) itself, not a dummy variable. So \( g(x) = \int_2^x x^2 \sin x \, dx \) is actually \( g(x) = \int_2^x f(x) \, dx \) where the integrand is a function of \( x \), not \( t \). 4. **Derivative:** Since the upper limit is \( x \), and the integrand depends on \( x \), the derivative is $$ g'(x) = \frac{d}{dx} \int_2^x x^2 \sin x \, dt = x^2 \sin x \cdot \frac{d}{dx} x + \int_2^x \frac{d}{dx} (x^2 \sin x) \, dt $$ But since the integral is with respect to \( t \), and the integrand depends on \( x \), the derivative is $$ g'(x) = x^2 \sin x + \int_2^x \frac{d}{dx} (x^2 \sin x) \, dt $$ However, this is unusual and likely a typo. Usually, the integrand is a function of \( t \), not \( x \). Assuming the problem meant \( g(x) = \int_2^x t^2 \sin t \, dt \), then by the Fundamental Theorem of Calculus, $$ g'(x) = x^2 \sin x $$ **Final answer:** \( g'(x) = x^2 \sin x \).