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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    \[\int_{-1}^{1}{({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})({{e}^{x}}-{{e}^{-x}})}dx\]is equal to

    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,       \[f(-x)=-f(x)\] \[\therefore \]  \[\int_{-1}^{1}{f(x)}dx=0\] \[\therefore \]  \[\int_{-1}^{1}{({{e}^{{{x}^{3}}}}+{{e}^{-{{x}^{3}}}})}({{e}^{x}}-{{e}^{-x}})dx=0\]


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