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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[{{I}_{n}}=\int_{0}^{\infty }{{{e}^{-x}}{{x}^{n-1}}dx,}\] then \[\int_{0}^{\infty }{{{e}^{-\lambda x}}{{x}^{n-1}}dx=}\]

    A) \[\lambda {{I}_{n}}\]        

    B) \[\frac{1}{\lambda }{{I}_{n}}\]

    C) \[\frac{{{I}_{n}}}{{{\lambda }^{n}}}\]                            

    D) \[{{\lambda }^{n}}{{I}_{n}}\]

    Correct Answer: C

    Solution :

    • Putting \[\lambda x=t,\lambda dx=dt\]                   
    • we get , \[\int_{0}^{\infty }{{{e}^{-\lambda x}}{{x}^{n-1}}dx}\]                   
    • \[=\frac{1}{{{\lambda }^{n}}}\int_{0}^{\infty }{{{e}^{-t}}{{t}^{n-1}}}dt\]\[=\frac{1}{{{\lambda }^{n}}}\int_{0}^{\infty }{{{e}^{-x}}{{x}^{n-1}}dx=\frac{{{I}_{n}}}{{{\lambda }^{n}}}}\].


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