A) \[\frac{(e-1)}{2}\]
B) \[{{e}^{2}}-1\]
C) \[2(e-1)\]
D) \[e-1\]
Correct Answer: A
Solution :
[a] Let \[I=\int\limits_{0}^{1}{x{{e}^{{{x}^{2}}dx}}}\] Let \[{{x}^{2}}=t\] \[\Rightarrow 2x\,\,dx=dt\] \[\Rightarrow xdx=\frac{dt}{2}\] when \[x=0,t=0\] then \[x=1,t=1\] \[x=1,t=1\] \[\Rightarrow I=\frac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\frac{1}{2}\left[ {{e}^{t}} \right]_{0}^{1}\] \[=\frac{1}{2}\left[ {{e}^{{{x}^{2}}}} \right]_{0}^{1}=\frac{1}{2}[e-{{e}^{0}}]=\frac{e-1}{2}\]
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