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Math Help - Trapezium Rule

  1. #1
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    Trapezium Rule

    Hi,

    I need help checking these, please.

    i.) Use the trapezium rule with intervals of width 0.5 to find an approximation for \int_1^{2.5}  \frac{1}{1\,+\,\ln x}\,dx


    h = 0.5

    \begin{vmatrix} x_r & 1 & 1.5 & 2.0 & 2.5 \\y_r & 1 & 0.7115 & 0.5906 & 0.5218 \end{vmatrix}

    \int_1^{2.5}  \frac{1}{1\,+\,\ln x}dx = \frac{1}{2}\cdot0.5\{1+2(0.7115+0.5906)\,+\,0.5218  \}


    \approx 1.0315


    ii.) Use the trapezium rule with intervals of width 0.4 to find an approximate value for \int_0^{1.6} xe^{x^2}\,dx

    Calculate the exact value of \int_0^{1.6} xe^{x^2}\,dx leaving your answer in terms of e.


    \int_0^{1.6} xe^{x^2} dx

    h = 0.4

    \begin{vmatrix} x_r & 0 & 0.4 & 0.8 & 1.2 & 1.6  \\y_r & 0 & 0.4694 & 1.517 & 5.065 & 20.70  \end{vmatrix}

    \int_0^{1.6}  xe^{x^2} dx = \frac{1}{2}\cdot0.4\{2(0.4694+1.517+5.065)+20.70\}

    \approx 6.961


    \int_0^{1.6} xe^{x^2} dx

    \approx 5.96791 How can this be expressed in terms of e?

    Last edited by Hellbent; November 9th 2010 at 04:15 PM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    1. Okay.

    2. Right.

    For this, I'll use a substitution of x^2 = u

    du = 2x dx

    dx = 1/2x du

    u = 1.6^2 = 2.56
    u = 0^2 = 0

    \begin{array}{rcl}<br />
\displaystyle \int^{1.6}_0 xe^{x^2}\ dx &=&\displaystyle \int^{2.56}_0 \frac{xe^u}{2x}\ du \\<br />
&& \\<br />
&=&\displaystyle \dfrac12 \int^{2.56}_0 e^u\ du \\<br />
&& \\<br />
&=& \dfrac12 \left[e^u\right]^{2.56}_0  \\<br />
&& \\<br />
&=& \dfrac12 \left[e^{2.56} - 1\right] \end{array}
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