Let f,g:(0,) R be two functions defined by f(x)= _-x^x(|t|-t^2)e^-t^2dt and g(x)= _0^x^2t^…

Question

Let $f,g:(0,\infty)	o R$ be two functions defined by $f(x)=\int _{-x}^{x}(|t|-{t}^{2}){e}^{-{t}^{2}}dt$ and $g(x)=\int _{0}^{{x}^{2}}{t}^{1/2}{e}^{-t}dt$. Then, the value of $9(f(\sqrt{{\log }_{e}9})+g(\sqrt{{\log }_{e}9}))$ is equal to :

Accepted Answer

$\begin{gathered} f(x)=\int _{-x}^{x}(|t|-{t}^{2}){e}^{-{t}^{2}}dt \ \Rightarrow {f}^{'}(x)=2\cdot (|x|-{x}^{2}){e}^{-{x}^{2}}\ldots \ldots \ldots ..(1) \ g(x)=\int _{0}^{{x}^{2}}{t}^{\frac{1}{2}}{e}^{-t}dt \ {g}^{'}(x)={xe}^{-{x}^{2}}(2x)-0 \ {f}^{'}(x)+{g}^{'}(x)=2{xe}^{-{x}^{2}}-2{x}^{2}{e}^{-{x}^{2}}+2{x}^{2}{e}^{-{x}^{2}} \end{gathered}$

Integrating both sides w.r.t. $x$

$\begin{gathered} f(x)+g(x)=\int _{0}^{\alpha }2{xe}^{-{x}^{2}}dx \ {x}^{2}=t \ \Rightarrow {\int }_{\sqrt{\alpha }}^{0}{e}^{-t}dt={[-{e}^{-t}]}_{\sqrt{\alpha }}^{0} \ =-{e}^{({\log }_{c}(9{)}^{-1})+1} \ \Rightarrow 9(f(x)+g(x))=(1-\frac{1}{9})9=8 \end{gathered}$

Solutions · 1