let f(x) be a twice - differentiable function with selected values of f(x) and its derivatives shown below…

Explanation:

Step1: Use integration - by - parts

Let $u=-2x$ and $dv = f''(x)dx$. Then $du=-2dx$ and $v = f'(x)$. By the integration - by - parts formula $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$, we have $\int_{1}^{2}-2xf''(x)dx=\left[-2xf'(x)\right]{1}^{2}+2\int{1}^{2}f'(x)dx$.

Step2: Evaluate $\left[-2xf'(x)\right]_{1}^{2}$

$\left[-2xf'(x)\right]_{1}^{2}=-2\times2\times f'(2)-(-2\times1\times f'(1))=-4f'(2) + 2f'(1)$. Substitute $f'(1)=-1$ and $f'(2)=3$: $-4\times3+2\times(-1)=-12 - 2=-14$.

Step3: Evaluate $2\int_{1}^{2}f'(x)dx$

By the fundamental theorem of calculus, $2\int_{1}^{2}f'(x)dx=2\left[f(x)\right]_{1}^{2}=2(f(2)-f(1))$. Substitute $f(1)=-3$ and $f(2)=-1$: $2(-1-(-3))=2\times2 = 4$.

Step4: Combine the results

$\int_{1}^{2}-2xf''(x)dx=-14 + 4=-10$.

Answer:

$-10$