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Math Help - Trapezium Rule

  1. #1
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    Trapezium Rule

    Hi,

    A little help required and some checking.

    The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for \int_0^1 e^{-x}\,dx.
    Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.

    My graphs are different from Wolfram's. Here they are: e^-x and -e^-x
    EXPLANATION PART: I don't know this part. Help required.

    Calculate the approximate and exact values respectively, to 3 d.p.

    h = \frac{1-0}{2} = 0.5

    \begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}

    \int_0^1 e^{-x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)<br />
= 0.645

    \int_0^1 e^{-x}\,dx = \bigg[-e^{-x}\bigg]_0^1

    =[-0.368] - [-1]

    = 0.632

    Another approximation to \int_0^1 e^{-x}, is to be calculated by using two trapezia of unequal width. The first trapezium has width h and the second has width (1 - h), so that the three ordinates are at x = 0, x = h and x = 1. Show that the total area T of these two trapezia is given by T = \frac{1}{2}(e^{-1} + h(1 - e^{-1})+e^{-k}).

    Width = h and (1 - h)

    x = 0, x = h, x = 1

    T=\frac{1}{2}h(e^0+e^{-h})+(1-h)(e^{-h}+e^{-1})

    =\frac{1}{2}{h+e^{-h}+e^{-1} - he^{-1}}

    T= \frac{1}{2}{e^{-1}+h(1-e^{-1})+e^{-h}}


    Show that the value of h for which T is a minimum is given by h = \ln\bigg(\frac{e}{e-1}\bigg).

    \frac{dT}{dh} = \frac{1}{2}(1-e^{-1})-e^{-h}

    0 = 1 - e^{-1} - e^{-h}

    -1 + e^{-1} = -e^{-h}

    -\ln\,1- \frac{1}{e} = h

    -\ln\,\frac{e-1}{e} = h

    \ln\bigg(\frac{e}{e-1}\bigg) = h
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    A little help required and some checking.

    The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for \int_0^1 e^{-x}\,dx.
    Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.

    My graphs are different from Wolfram's. Here they are: e^-x and -e^-x
    EXPLANATION PART: I don't know this part. Help required.

    Calculate the approximate and exact values respectively, to 3 d.p.

    h = \frac{1-0}{2} = 0.5

    \begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}

    \int_0^1 e^{-x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)<br />
= 0.645

    \int_0^1 e^{-x}\,dx = \bigg[-e^{-x}\bigg]_0^1

    =[-0.368] - [-1]

    = 0.632

    Another approximation to \int_0^1 e^{-x}, is to be calculated by using two trapezia of unequal width. The first trapezium has width h and the second has width (1 - h), so that the three ordinates are at x = 0, x = h and x = 1. Show that the total area T of these two trapezia is given by T = \frac{1}{2}(e^{-1} + h(1 - e^{-1})+e^{-k}).

    Width = h and (1 - h)

    x = 0, x = h, x = 1

    T=\frac{1}{2}h(e^0+e^{-h})+(1-h)(e^{-h}+e^{-1})

    =\frac{1}{2}{h+e^{-h}+e^{-1} - he^{-1}}

    T= \frac{1}{2}{e^{-1}+h(1-e^{-1})+e^{-h}}


    Show that the value of h for which T is a minimum is given by h = \ln\bigg(\frac{e}{e-1}\bigg).

    \frac{dT}{dh} = \frac{1}{2}(1-e^{-1})-e^{-h}

    0 = 1 - e^{-1} - e^{-h}

    -1 + e^{-1} = -e^{-h}

    -\ln\,1- \frac{1}{e} = h

    -\ln\,\frac{e-1}{e} = h

    \ln\bigg(\frac{e}{e-1}\bigg) = h
    1. Have you tried drawing the graph? I would say it's because the function \displaystyle e^{-x} is convex (check by showing that the second derivative is always positive), so all chords will lie above the graph.
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  3. #3
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    Yes, I did draw the graphs. The ones I gave are Wolfram's.
    The graphs I drew are probably okay, so forget that part. Hehe
    Last edited by Hellbent; November 14th 2010 at 10:24 AM.
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  4. #4
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    Delete.
    Last edited by Hellbent; November 14th 2010 at 10:23 AM.
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  5. #5
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    Is the rest of what I did okay?
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