Let g(x) be a non-constant twice differentiable function defined such that y=g(x) is symmetric about the line x=2, ${\mathrm{I}}_1=\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\mathrm{g}(\mathrm{x}+2)\mathrm{sinx}\mathrm{dx}\mathrm{and}{\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+{\mathrm{e}}^{\mathrm{g}\text{'}\left(\mathrm{x}\right)}}\mathrm{dx},$ then which of the following is incorrect?

Here, g(2-x) =g(2+x)
Hence, g(2+x) sin x is an odd function $\therefore {\mathrm{I}}_1=0$
$\mathrm{Now},\mathrm{g}(2-(2-\mathrm{x}))=\mathrm{g}(2+2-\mathrm{x})\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}(4-\mathrm{x})$

$\therefore {\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)=-{\mathrm{g}}^{\text{'}}(4-\mathrm{x})$

$\mathrm{Also},{\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+{\mathrm{e}}^{{\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)}}\mathrm{dx}.....\left(\mathrm{i}\right)\Rightarrow {\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+{\mathrm{e}}^{{\mathrm{g}}^{\text{'}}(4-\mathrm{x})}}\mathrm{dx}$

$\Rightarrow {\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+{\mathrm{e}}^{-{\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)}}\mathrm{dx}\Rightarrow {\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}\frac{{\mathrm{e}}^{{\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)}}{{\mathrm{e}}^{{\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)}+1}\mathrm{dx}....\left(\mathrm{ii}\right)$

$$\begin{gathered} \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} 2{\mathrm{I}}_2=\underset{0}{\overset{4}{\int }}1\mathrm{dx} \\ \therefore {\mathrm{I}}_2=2 \end{gathered}$$