2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    Let\[{{I}_{1}}=\int_{\alpha }^{\pi -\alpha }{xf}(\sin x)\,\,dx,\,\,{{I}_{2}}\]\[=\int_{xf(\sin x)\,\,dx}^{\pi -\alpha }{f(\sin x)}\,\,dx\], then \[{{I}_{2}}\] is equal to

    A) \[\frac{\pi }{2}{{I}_{1}}\]                                              

    B) \[\pi {{I}_{1}}\]

    C) \[\frac{2}{\pi }{{I}_{1}}\]                              

    D)        \[2{{I}_{1}}\]

    Correct Answer: C

    Solution :

    Since,\[{{I}_{1}}=\int_{a}^{\pi -a}{x\,\,f(\sin x)\,\,dx}\]                 \[=\int_{a}^{\pi -a}{(\pi -x)f(\sin (\pi -x))}\,\,dx\]                 \[=\int_{a}^{\pi -a}{(\pi -x)f(\sin x)}\,\,dx\]                 \[=\int_{a}^{\pi -a}{\pi f(\sin x)}\,\,dx-{{I}_{1}}\] \[\Rightarrow \]               \[2{{I}_{1}}={{I}_{2}}\] \[\Rightarrow \]               \[{{I}_{2}}=\frac{2}{\pi }{{I}_{1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner