Question

How to solve this. I'm feeling lazy

Answer 1

`\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}`


`\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]`

now, apply the bounds to
`\bf \left[\cfrac{}{} 9s \right]_3^8` to get the definite integral value, and simplify away