A) \[\frac{{{e}^{2}}}{2}-2e\]
B) \[{{e}^{2}}-2e\]
C) \[2({{e}^{2}}-e)\]
D) \[2{{e}^{2}}-2e\]
E) \[0\]
Correct Answer: E
Solution :
Let \[I=\int{\underset{I}{\mathop{({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})}}\,}\underset{II}{\mathop{({{e}^{x}}-{{e}^{-x}})dx}}\,\] \[I=({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})\] \[-\int{(3{{x}^{2}}{{e}^{{{x}^{3}}}}-3{{x}^{2}}{{e}^{-{{x}^{3}}}})}({{e}^{x}}+{{e}^{-x}})dx\] \[I=({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})\] \[-3{{x}^{2}}[({{e}^{{{x}^{3}}}}-{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})\] \[-\int{3{{x}^{2}}({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})}dx]\] \[I=({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})\] \[-3{{x}^{2}}({{e}^{{{x}^{3}}}}-{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})+9{{x}^{4}}I\] \[\Rightarrow \] \[I=(1-9{{x}^{4}})=({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})-3{{x}^{2}}\] \[({{e}^{{{x}^{3}}}}-{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})\] \[\therefore \] \[I=\frac{\begin{align} & ({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})-3{{x}^{2}}({{e}^{{{x}^{3}}}}-{{e}^{-{{x}^{3}}}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{e}^{x}}-{{e}^{-x}}) \\ \end{align}}{1-9{{x}^{4}}}\] \[\therefore \] \[\int_{-1}^{1}{({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})}({{e}^{x}}-{{e}^{-x}})dx\] \[=\left[ \frac{\begin{align} & ({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}+{{e}^{-x}})-3{{x}^{2}}({{e}^{{{x}^{3}}}}-{{e}^{-{{x}^{3}}}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{e}^{x}}-{{e}^{-x}}) \\ \end{align}}{1-9{{x}^{4}}} \right]_{-1}^{1}\] \[=\frac{(e+{{e}^{-1}})(e+{{e}^{-1}})-3(e-{{e}^{-1}})(e-{{e}^{-1}})}{1-9}\] \[-\frac{({{e}^{-1}}+e)({{e}^{-1}}+e)-3({{e}^{-1}}-e)({{e}^{-1}}-e)}{1-9}\] \[=0\] Alternate Let \[f(x)=({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})\] This is an odd function Since, \[f(-x)=-f(x)\] \[\therefore \] \[\int_{-1}^{1}{f(x)}dx=0\] \[\therefore \] \[\int_{-1}^{1}{({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})}({{e}^{x}}-{{e}^{-x}})dx=0\]You need to login to perform this action.
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