A) \[{{e}^{\frac{-{{x}^{2}}}{2}}}\]
B) \[{{e}^{\frac{{{x}^{2}}}{4}}}\]
C) \[{{e}^{\frac{-{{x}^{2}}}{4}}}\]
D) \[{{e}^{\frac{{{x}^{2}}}{4}}}\]
Correct Answer: B
Solution :
\[{{I}_{1}}=\int\limits_{0}^{x}{{{e}^{t(x-t)}}dt}\] \[t(x-t)=-[{{t}^{2}}-tx]\,=-\,\left[ {{\left( t-\frac{x}{2} \right)}^{2}}-\frac{{{x}^{2}}}{4} \right]\] \[=\frac{{{x}^{2}}}{4}\,-{{\left( t-\frac{x}{2} \right)}^{2}}\] \[{{I}_{1}}=\int\limits_{0}^{x}{{{e}^{\frac{{{x}^{2}}}{4}{{\left( t-\frac{x}{2} \right)}^{2}}}}dt={{e}^{\frac{{{x}^{2}}}{4}}}\int\limits_{0}^{x}{{{e}^{\frac{-{{(2t-x)}^{2}}}{4}}}}}\] \[dt;\,2t-x=y\] \[\Rightarrow \,\,dt=\frac{dy}{2}\]\[=\frac{{{e}^{\frac{{{x}^{2}}}{4}}}}{2}\,\int\limits_{-x}^{x}{{{e}^{\frac{-{{y}^{2}}}{4}dy}}}\] \[{{I}_{1}}={{e}^{\frac{{{x}^{2}}}{4}}}\,\int\limits_{0}^{x}{\frac{-{{t}^{2}}}{4}dt\,={{e}^{\frac{{{x}^{2}}}{4}}}.\,{{I}_{2}}\,\Rightarrow \,\,\frac{{{I}_{1}}}{{{I}_{2}}}\,={{e}^{\frac{{{x}^{2}}}{4}}}}\]You need to login to perform this action.
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