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anonymous

  • one year ago

please helpp.. question in comments

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  1. anonymous
    • one year ago
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  2. mathmate
    • one year ago
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    To solve a problem of integration, it is best to make a plot of the function to be integrated in order to detect anomalies which can cause problems, for example, vertical asymptotes, discontinuities, etc. Sometimes we will also find symmetries, which will halve our efforts by doing only half of the interval, or anti-symmetries, in which case we just write down the answer. In this case, a sketch of the functions looks like the following, so we need to do real work. |dw:1443202170243:dw| We can also estimate the answer as approximately 2/3 of the area of the rectangle, or (2/3)*0.7*pi/2 = 0.73 The trapezium rule is given by \(\int_a^b f(x)dx = \frac{h}{2}(f(a)+2f(a+h)+2f(a+2h)....2f(b-h)+f(b) )\) where h=(b-a)/k In the given problem, k=3 (intervals), so h=(pi/2)/3=pi/6 a=0, b=pi/2. You will need to evaluate f(x)=ln(1+sin(x)) for x=0, pi/6, pi/3, pi/2 and evaluate the integral accordingly. As you probably know, the more intervals you use, the better the results will be. For more information and reading, try: http://www.mathwords.com/t/trapezoid_rule.htm If you wish, you could post your results for checking.

  3. anonymous
    • one year ago
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    thank you

  4. mathmate
    • one year ago
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    You're welcome! :)

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