7. -/6.25 points details my notes suppose that f(1) = 2, f(4) = 9, f (1) = 7, f (4) = 4, and f is…

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = x$ and $\mathrm{d}v=f''(x)\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$ and $v = f'(x)$. So, $\int_{1}^{4}xf''(x)\mathrm{d}x=[xf'(x)]{1}^{4}-\int{1}^{4}f'(x)\mathrm{d}x$.

Step2: Evaluate $[xf'(x)]_{1}^{4}$

$[xf'(x)]_{1}^{4}=4f'(4)-1\times f'(1)$. Given $f'(1) = 7$ and $f'(4)=4$, then $4f'(4)-f'(1)=4\times4 - 7=16 - 7 = 9$.

Step3: Evaluate $\int_{1}^{4}f'(x)\mathrm{d}x$

By the fundamental theorem of calculus, $\int_{1}^{4}f'(x)\mathrm{d}x=f(4)-f(1)$. Given $f(1) = 2$ and $f(4)=9$, then $f(4)-f(1)=9 - 2=7$.

Step4: Calculate the final result

$\int_{1}^{4}xf''(x)\mathrm{d}x=[xf'(x)]{1}^{4}-\int{1}^{4}f'(x)\mathrm{d}x=9-7 = 2$.

Answer:

$2$