Question

If $\underset{1}{\overset{3}{\int }}{x}^2{F}^{\prime }\left(x\right)dx=-12$ and $\underset{1}{\overset{3}{\int }}{x}^3{F}^{\prime \prime }\left(x\right)dx=40$, then the correct explanation(s) is(are)

• A. $\underset{1}{\overset{3}{\int }}f\left(x\right)dx=-12$
• B. $9f^{\prime }\left(3\right)+f^{\prime }\left(1\right)-32=0$
• C. $\underset{1}{\overset{3}{\int }}f\left(x\right)dx=12$
• D. $9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$

Answer 1

Answer: (A), (D)

$9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$

$\underset{1}{\overset{3}{\int }}{x}^2{F}^{\prime }\left(x\right)dx=-12$ and $\underset{1}{\overset{3}{\int }}{x}^3{F}^{\prime \prime }\left(x\right)dx=40$
$\underset{1}{\overset{3}{\int }}{x}^3{F}^{\prime \prime }\left(x\right)=x^3{F}^{\prime }\left(x\right){]}_1^3-\underset{1}{\overset{3}{\int }}3x^2{F}^{\prime }\left(x\right)$
$40=27F^{\prime }\left(3\right)-F^{\prime }\left(1\right)+36$
$4=9\left(f^{\prime }\right(3)+4)-f^{\prime }\left(1\right)$
$\Rightarrow 9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$
Also $\underset{1}{\overset{3}{\int }}{x}^2{F}^{\prime }\left(x\right)dx$
$=x^{2}F\left(x\right){]}_1^3-\underset{1}{\overset{3}{\int }}2xF\left(x\right)dx=-12$
$9F\left(3\right)-F\left(1\right)-2\underset{1}{\overset{3}{\int }}xF\left(x\right)dx=-12$
$-36-0-2\underset{1}{\overset{3}{\int }}f\left(x\right)dx=-12$
$-36+12=2\underset{1}{\overset{3}{\int }}f\left(x\right)dx$
$-12=\underset{1}{\overset{3}{\int }}f\left(x\right)dx$