Suppose that $f$ is a continuous function such that $f(3)=8$ and $f(9)=6$. If $\mathrm{F}(\mathrm{x})$ is defined for each real number $x$ by
\[
F(x)=\int_{x}^{x^{2}} f(t) d t
\]
then what is the value of $F^{\prime}(3)$ ?
Answer
So, the value of \(F'(3)\) is \(\boxed{28}\).
Step 1 ：Given the function \(F(x) = \int_{x}^{x^{2}} f(t) dt\), we want to find \(F'(3)\), which is the derivative of \(F\) at \(x = 3\).
Step 2 ：By the chain rule and the Fundamental Theorem of Calculus, we have \(F'(x) = f(x^2) * 2x - f(x)\).
Step 3 ：So, \(F'(3) = f(3^2) * 2*3 - f(3)\).
Step 4 ：Substituting the given values \(f(3) = 8\) and \(f(9) = 6\), we get \(F'(3) = 6 * 2*3 - 8\).
Step 5 ：Solving the above expression, we get \(F'(3) = 36 - 8 = 28\).
Step 6 ：So, the value of \(F'(3)\) is \(\boxed{28}\).
