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Thread: 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 $\displaystyle 0.5$ to find an approximation for $\displaystyle \int_1^{2.5} \frac{1}{1\,+\,\ln x}\,dx$


    $\displaystyle h = 0.5$

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

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


    $\displaystyle \approx 1.0315$


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

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


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

    $\displaystyle h = 0.4$

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

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

    $\displaystyle \approx 6.961$


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

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

    Last edited by Hellbent; Nov 9th 2010 at 03:15 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

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