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    Use the trapezium rule with four ordinates (three strips) to find an approximate value for?
    Integral (limits 3-0) sqrt(2^x) dx

    (How do you find the width of first and second strip, 1st height, sum of mid heights and find the last height, with no diagram)

    Also when would you sub the limits into the integral and then minus them?

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    Well, the integral is between 3 and 0, so the strips are going to be between these values of x, so the strips will be 0-1,1-2,2-3, each with a width of 1. The height will be the output of f(x)=sqrt(2^x) for the first x value of each strip, ie h_1=sqrt(2^0)
    I don't know what you mean for the sum of mid heights, and the last height you can find the same way as the first.
    I'd assume then you find the area of each strip and sum them to get your approximation.
    There is a formula for the trapezium rule though.
    Something like 1/2h(h_0 +h_n 2(h_1 + ... + h_n-1))..
    (i can;'t remember exactly)
    Though you could just integrate that one.

    I think that makes sense, though I've already made some pretty stupid errors today, so I may have done that again.

     \int^3_0{\sqrt{2^x}} dx
     2^x=e^{x \ln2}
     u=e^{x \ln2}, \frac{dx}{du}=\frac{1}{\ln2 e^{x \ln2}}
     \frac{1}{ln2} \int^{2^3}_{1}{u^{\frac{1}{2}} u^{-1}} du
     \frac{1}{ln2}[2u^{\frac{1}{2}}]{_{1}^{2^3}} \approx 5.28
    I think that's right, so your answer should be something close to that.
 
 
 
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Updated: September 27, 2010

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