show instructions question 8 question 8 of the table above gives selected values of twice differentiable…

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 $u = f(x)$ and $\mathrm{d}v=g''(x)\mathrm{d}x$. Then $\mathrm{d}u = f'(x)\mathrm{d}x$ and $v = g'(x)$. So, $\int_{2}^{4}f(x)g''(x)\mathrm{d}x=f(x)g'(x)|{2}^{4}-\int{2}^{4}g'(x)f'(x)\mathrm{d}x$.

Step2: Evaluate $f(x)g'(x)|_{2}^{4}$

$f(x)g'(x)|_{2}^{4}=f(4)g'(4)-f(2)g'(2)$. From the table, $f(2) = 7$, $f(4)=13$, $g'(2) = 1$, $g'(4)=7$. So, $f(4)g'(4)-f(2)g'(2)=13\times7 - 7\times1=91 - 7=84$.

Step3: Evaluate $\int_{2}^{4}g'(x)f'(x)\mathrm{d}x$

Since $f'(x)=3$ for all $x$, then $\int_{2}^{4}g'(x)f'(x)\mathrm{d}x = 3\int_{2}^{4}g'(x)\mathrm{d}x$. By the fundamental theorem of calculus, $\int_{2}^{4}g'(x)\mathrm{d}x=g(4)-g(2)$. From the table, $g(2) = 2$ and $g(4)=9$. So, $\int_{2}^{4}g'(x)\mathrm{d}x=9 - 2 = 7$. Then $3\int_{2}^{4}g'(x)\mathrm{d}x=3\times7 = 21$.

Step4: Calculate the value of $\int_{2}^{4}f(x)g''(x)\mathrm{d}x$

$\int_{2}^{4}f(x)g''(x)\mathrm{d}x=f(x)g'(x)|{2}^{4}-\int{2}^{4}g'(x)f'(x)\mathrm{d}x$. Substitute the values we found: $\int_{2}^{4}f(x)g''(x)\mathrm{d}x=84-21 = 63$.

Answer:

d. 63