Question

If $F\left(x\right)={\int }_1^{x}f\left(t\right)dt$, where $f\left(t\right)={\int }_1^{t^2}\frac{\sqrt{1+u^4}}{u}du$, then the value of $F^{\prime \prime }\left(2\right)$ equals to

• A. $\frac{7}{4\sqrt{17}}$
• B. $\frac{15}{\sqrt{17}}$
• C. $\sqrt{257}$
• D. $\frac{15\sqrt{17}}{68}$

Answer 1

The correct option is C $\sqrt{257}$

$f\left(t\right)={\int }_1^{t^2}\frac{\sqrt{1+u^4}}{u}du$
Applying Leibtnitz rule to differentiate under integration, we get
$\Rightarrow f^{\prime }\left(t\right)=\frac{\sqrt{1+t^8}}{t^2}\frac{d\left(t^2\right)}{dt}-\sqrt{2}\frac{d\left(1\right)}{dt}$

$\Rightarrow f^{\prime }\left(t\right)=\frac{2\sqrt{1+t^8}}{t}$
$F\left(x\right)={\int }_1^{x}f\left(t\right)dt$
$\Rightarrow F^{\prime }\left(x\right)=f\left(x\right)$
$\Rightarrow F^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)$
$\Rightarrow F^{\prime \prime }\left(2\right)=\frac{2\sqrt{257}}{2}=\sqrt{257}$
Hence, option 'C' is correct.