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JEE Main & Advanced Mathematics Definite Integration Question Bank Fundamental definite integration, Definite integration by substitution

  • question_answer
    \[\int_{0}^{1}{\frac{dx}{{{[ax+b(1-x)]}^{2}}}}=\]                                     [SCRA 1986]

    A)                 \[\frac{a}{b}\]  

    B)                 \[\frac{b}{a}\]

    C)                 \[a\,b\]

    D)                 \[\frac{1}{a\,b}\]

    Correct Answer: D

    Solution :

               Let \[I=\int_{0}^{1}{\frac{dx}{{{[(a-b)x+b]}^{2}}}}\]            Put \[t=(a-b)x+b\Rightarrow dt=(a-b)dx\]            As \[x=1\Rightarrow t=a\]and \[x=0\Rightarrow t=b\], then                      \[I=\frac{1}{a-b}\int_{b}^{a}{\frac{1}{{{t}^{2}}}}dt=\frac{1}{(a-b)}\left[ -\frac{1}{t} \right]_{b}^{a}=\frac{1}{(a-b)}\left( \frac{a-b}{ab} \right)=\frac{1}{ab}\].


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