JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

  • question_answer Area bounded by the curve \[y=x{{e}^{{{x}^{2}}}},\] \[x-\]axis and the ordinates \[x=0,\,\,x=a\]

    A)            \[\frac{{{e}^{{{a}^{2}}}}+1}{2}\]sq. unit                       

    B)            \[\frac{{{e}^{{{a}^{2}}}}-1}{2}\]sq. unit

    C)            \[{{e}^{{{a}^{2}}}}+1\]sq. unit                                          

    D)            \[{{e}^{{{a}^{2}}}}-1\]sq. unit

    Correct Answer: B

    Solution :

                       Required area is \[\int_{0}^{a}{y\,\,dx=\int_{0}^{a}{x{{e}^{{{x}^{2}}}}dx}}\]                    We put \[{{x}^{2}}=t\Rightarrow dx=\frac{dt}{2x}\] as \[x=0\Rightarrow t=0\] and \[x=a\Rightarrow t={{a}^{2}}\], then it reduces to                    \[\frac{1}{2}\int_{0}^{{{a}^{2}}}{{{e}^{t}}dt=\frac{1}{2}[{{e}^{t}}]_{0}^{{{a}^{2}}}=\frac{{{e}^{{{a}^{2}}}}-1}{2}}\] sq. unit.

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