A) \[{{e}^{2}}-1\]
B) \[2\]
C) \[\frac{{{e}^{2}}-1}{2}\]
D) \[\frac{{{e}^{2}}-1}{4}\]
Correct Answer: D
Solution :
[d] \[\int\limits_{0}^{1}{\underset{n\,\to \,\infty }{\mathop{\lim }}\,}\sum\limits_{k=0}^{n}{\frac{{{x}^{k+2}}2k}{k!}}dx\]=\[\int\limits_{0}^{1}{{{x}^{2}}\,\,\underset{n\,\to \,0}{\mathop{\lim }}\,\sum\limits_{k=0}^{n}{\frac{{{\left( 2x \right)}^{k}}}{k!}dx=\int\limits_{{}}^{1}{{{x}^{2}}{{e}^{2x}}dx}}}\]\[=\left. \frac{{{x}^{2}}{{e}^{2x}}}{2} \right|_{0}^{1}-\int\limits_{0}^{1}{x{{e}^{2x}}dx}\]\[=\frac{{{e}^{2}}}{2}\left[ \frac{x{{e}^{2x}}}{2}-\frac{{{e}^{2x}}}{4} \right]_{0}^{1}\]\[=\frac{{{e}^{2}}}{2}-\left[ \left( \frac{{{e}^{2}}}{2}-\frac{{{e}^{2}}}{4} \right)-\left( -\frac{1}{4} \right) \right]=\frac{{{e}^{2}}-1}{4}\]
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