1. Find the derivative of the function (a) $g(x) = \int_2^x t^2 \sin t \, dt$. Step 1: State the problem. We want to find $g'(x)$ where $g(x) = \int_2^x t^2 \sin t \, dt$. Step 2: Use the Fundamental Theorem of Calculus. If $G(x) = \int_a^x f(t) \, dt$, then $G'(x) = f(x)$. Step 3: Apply the theorem. Here, $f(t) = t^2 \sin t$, so $$g'(x) = x^2 \sin x.$$ Step 4: Explanation. The derivative of an integral with a variable upper limit is just the integrand evaluated at that upper limit. Final answer: $$g'(x) = x^2 \sin x.$$