![\bf g(x)=x^2\int\limits_{3}^{x}[4f(s)-9]ds\\\\ -----------------------------\\\\ \textit{let us use the product rule to get g'(x)} \\\\ \cfrac{dg}{dx}\implies 2x\left[ \int\limits_{3}^{x}[4f(s)-9]ds \right]\quad +\quad x^2[4f(x)-9] \\\\ g'(8)\implies \begin{array}{llll} 2(8)\left[ \int\limits_{3}^{8}[4f(s)-9]ds \right]\quad +\quad (8)^2[4f(8)-9] \\\\ 2(8)\left[ 4\underline{\int\limits_{3}^{8}f(s)ds} - \int\limits_{3}^{8} 9\cdot ds \right]\quad +\quad (8)^2[4\underline{f(8)}-9] \end{array} ](/tpl/images/0329/4990/5e0c0.png)
![\bf \textit{now, we know that}\implies \begin{cases} f(8)=5 \\\\ \int\limits_{3}^{8}f(s)ds=3 \end{cases}\\\\ -----------------------------\\\\ g'(8)\implies 2(8)\left[ 4\cdot \underline{3} - \left[\cfrac{}{} 9s \right]_3^8\right]\quad +\quad (8)^2[4\cdot \underline{5}-9]](/tpl/images/0329/4990/31a83.png)
![\bf \left[\cfrac{}{} 9s \right]_3^8](/tpl/images/0329/4990/64d74.png)
to get the definite integral value, and simplify awayHow to solve this. i'm feeling lazy
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