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Thread: changing the order of integration

  1. #1
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    changing the order of integration

    I need to get sin1(1-cos1) for the answer but I keep getting wrong answer. Please someone help me !!
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  2. #2
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    Hello, sawawak!

    I too can get only so far . . .


    $\displaystyle \int^1_0\int^1_x y^{-1}\sin y\cos\left(\frac{x}{y}\right)\,dy\,dx$

    I need to get: .$\displaystyle \sin1(1-\cos1)$ for the answer.
    The limits are: .$\displaystyle \begin{Bmatrix}y = 1 \\ y=x\end{Bmatrix}\quad\begin{Bmatrix}x=1 \\ x=0\end{Bmatrix}$

    The region looks like this:
    Code:
            |
           1+ - - - - - *
            |:::::::::* :
            |:::::::*   :
            |:::::*     :
            |:::*       :
            |:*         :
          - * - - - - - + - -
            |           1

    Reversing the order of integration, we have: .$\displaystyle \begin{Bmatrix}x=y\\ x=0 \end{Bmatrix}\quad\begin{Bmatrix}y=1 \\ y=0 \end{Bmatrix}$

    And we have: .$\displaystyle \int^1_0\underbrace{\int^y_0 y^{-1}\cdot\sin y\cdot\cos\left(\frac{x}{y}\right)\,dx}\,dy$

    . . . . . . . . . . . . $\displaystyle y^{-1}\cdot\sin y\cdot\frac{1}{y}\cdot\sin\left(\frac{x}{y}\right) \,\bigg]^y_0 $

    . . . . . . . . . $\displaystyle = \;\bigg[y^{-1}\sin y\cdot\frac{1}{y}\cdot\sin\left(\frac{y}{y}\right) \bigg] - \bigg[y^{-1}\cdot\sin y \cdot\frac{1}{y}\cdot\sin\left(\frac{0}{y}\right)\ bigg] $

    . . . . . . . . . $\displaystyle = \;\frac{1}{y}\cdot\sin y\cdot\frac{1}{y}\cdot\sin(1) $

    . . . . . . . . . $\displaystyle = \;\sin(1)\,\frac{\sin y}{y^2}$


    Now we must integrate: .$\displaystyle \sin(1)\int^1_0 \frac{\sin y}{y^2}\,dy $


    Did I make some egregious blunders?

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  3. #3
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    The $\displaystyle \frac{1}{y}$ get absorbed on integration. I'll pick it up here

    $\displaystyle \int^1_0\int^y_0 y^{-1}\cdot\sin y\cdot\cos\left(\frac{x}{y}\right)\,dx\,dy = \int_0^1 \sin y \left. \sin \frac{x}{y} \right|_0^y\, dy = \int_0^1 \sin y \sin 1\, dy $ $\displaystyle = - \sin 1 \cos y |_0^1 = \sin 1 ( 1 - \cos 1) $
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