Given $f\in C^7[a,b]$ $x_k=a+kh$ where $h=\frac{b-a}{n},k=-1,0,\ldots,n+1$
Find a way to approximate $f''(x_k)$ with $f(x_{k-1}),f(x_k),f(x_{k+1})$ and then find a way to approximate $\int_{a}^{b}f(x)f''(x)dx$
I've found that $f''(x_k)=1/h^2(f(x_{k-1})-2f(x_k)+f(x_{k+1}))+h^2/24(f^{(4)}(C_1)+f^{(4)}(C_2))$
But I didn't know how to solve the other part(I know simpson rule and trapeze rule)
If $g \in C^2([a,b],\mathbb{R})$, then by the trapezoidal rule we have $$ \int_a^b g(x)dx = \frac{h}{2} \sum_{j=0}^{n-1} \left[ g(x_j) + g(x_{j+1}) \right] + O(h^2), \quad h \rightarrow 0, \quad nh = b-a > 0.$$ If $f \in C^4(\mathbb{R},\mathbb{R})$, then by Taylor's formula we have $$ f''(x) = \frac{f(x+h) - 2f(x)+f(x-h)}{h^2} + O(h^2), \quad h \rightarrow 0, \quad h \not = 0.$$ We will now combine these results. If $f \in C^4(\mathbb{R},\mathbb{R})$, then $$ g(x) = f(x)f''(x) = f(x)\frac{f(x+h) - 2f(x)+f(x-h)}{h^2} + O(h^2), \quad h \rightarrow 0, \quad h \not = 0.$$ If $[a,b] \subset \mathbb{R}$, then we certainly have $$ \int_a^b f(x) f''(x) = \frac{h}{2} \sum_{j=0}^{n-1} \left[f(x_j) \frac{f(x_{j+1}) - 2f(x_j)+f(x_{j-1})}{h^2} + f(x_{j+1}) \frac{f(x_{j+2}) - 2f(x_{j+1})+f(x_{j})}{h^2} \right]+ O(h^2), \quad h \rightarrow 0, \quad nh = b-a.$$
There are some concerns though. The original problem statements does not stipulate that $f$ is defined outside $[a,b]$, but makes explicit references to the value of $f$ at $a-h$ and at $b+h$. I have circumvented this issue by assuming that $f \in C^4(\mathbb{R}, \mathbb{R})$. If $f$ is not defined outside $[a,b]$, then do the partial integration $$ \int_a^b f(x) f''(x) = [f(x)f'(x)]_a^b - \int_a^b f'(x)^2 dx$$ and use the midpoint method to evaluated the last integral exploiting that $$ f'(t) = \frac{f(t+h/2) - f(t-h/2)}{h} + O(h^2), \quad h \rightarrow 0, \quad t \pm h/2 \in [a,b].$$ Assuming that $f \in C^7$ is unnecessary for the portion of the problem that you have presented. It may be necessary for the rest of the problem.