Question

If ${\int }_1^3{x}^2{F}^{\prime }\left(x\right)dx=-12$ and ${\int }_1^3{x}^3{F}^{\prime \prime }\left(x\right)dx=40$, then the correct expression(s) is (are)

• A. $9f^{\prime }\left(3\right)+f^{\prime }\left(1\right)-32=0$
• B. ${\int }_1^{3}f\left(x\right)dx=12$
• C. $9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$
• D. ${\int }_1^{3}f\left(x\right)dx=-12$

Answer 1

Answer: (C), (D)

The correct options are
C $9f^{\prime }\left(3\right)-f^{\prime }\left(1\right)+32=0$
D ${\int }_1^{3}f\left(x\right)dx=-12$

${\int }_1^{3}f\left(x\right)dx={\int }_1^{3}xF\left(x\right)dx=\left(F\right(x).\frac{x^2}{2}{)}_1^3-{\int }_1^3\frac{x^2}{2}.F^{{}^{\prime }}\left(x\right)dx=\frac{9}{2}(-4)-\frac{1}{2}(-12)=-12$
$f^{{}^{\prime }}\left(3\right)=F\left(3\right)+3F^{{}^{\prime }}\left(3\right)-1)f^{{}^{\prime }}\left(1\right)=F^{{}^{\prime }}\left(1\right)-2)$

For option C, subsitute equation $1)$ and $2)$

$9f^{{}^{\prime }}\left(3\right)-f^{{}^{\prime }}\left(1\right)+32\Rightarrow 9F\left(3\right)+27F^{{}^{\prime }}\left(3\right)-F^{{}^{\prime }}\left(1\right)+32$

$27F^{{}^{\prime }}\left(3\right)-F^{{}^{\prime }}\left(1\right)-4=0$

On substituting the values of F(x) we get the value as zero