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

  • question_answer The area of the region bounded by \[y=\,\,|x-1|\] and \[y=1\] is                                                                       [IIT Screening 1994]

    A)            2    

    B)            1

    C)            \[\frac{1}{2}\]                      

    D)            None of these

    Correct Answer: B

    Solution :

               \[y=x-1,\]if \[x>1\]and \[y=-(x-1),\]if \[x<1\] Area \[=\int_{0}^{1}{(1-x)dx+\int_{1}^{2}{(x-1)dx}}=\left[ x-\frac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\]                            \[=\left[ 1-\frac{1}{2} \right]+\left[ -\left( \frac{1}{2}-1 \right) \right]\]\[=\frac{1}{2}+\frac{1}{2}=1\].

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