$${given}\:{that}\:\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt}=\mathrm{1} \\$$$${and}\:{g}\left({n},{u}\right)=\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\:\underset{−\infty} {\overset{+\infty} {\int}}\left({x}−{u}\right)^{{n}} {e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt} \\$$$${and}\:{f}\left({n},{u}\right)=\frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}\left({t}−{u}\right)^{{n}} {e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt} \\$$$${ii}.\:{evaluate}\:{f}\left(\mathrm{1},\mathrm{0}\right) \\$$$${iii}.\:{evaluate}\:{f}\left(\mathrm{2},\mathrm{0}\right) \\$$$${iv}.\:{evaluate}\:{f}\left(\mathrm{1},\mu\right) \\$$
$${f}\left({n},{u}\right)=\left({x}−{u}\right)^{{n}} \frac{\mathrm{1}}{\sigma\sqrt{\mathrm{2}\pi}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−\frac{\left({t}−\mu\right)^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dt}=\left({x}−{u}\right)^{{n}} \\$$$${x},\:{n}\:\mathrm{and}\:{u}\:\mathrm{are}\:\mathrm{constant}\:\mathrm{for}\:\mathrm{the}\:\mathrm{integration}. \\$$
$${f}\left(\mathrm{1},\mathrm{0}\right)={x} \\$$$${f}\left(\mathrm{2},\mathrm{0}\right)={x}^{\mathrm{2}} \\$$$${f}\left(\mathrm{1},\mu\right)=\left({x}−\mu\right) \\$$$${x},{n}\:\mathrm{and}\:\mu\:\mathrm{are}\:\mathrm{constants}\:\mathrm{in}\:\mathrm{the}\:\mathrm{integral}. \\$$