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Math Help - Integrate functions and use calculus to solve problems

  1. #1
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    Question Integrate functions and use calculus to solve problems

    Use a numerical method to find the area enclosed by the graph of y = x▓ sin x, the x axis and the lines x = 1 and x = 2.6
    Use an interval of 0.4
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  2. #2
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    Quote Originally Posted by norivea View Post
    Use a numerical method to find the area enclosed by the graph of y = x▓ sin x, the x axis and the lines x = 1 and x = 2.6
    Use an interval of 0.4
    We can approximate via trapezoid rule for the function is continous.

    In order to use the trapezoid rule tou need to find the number of intervals.

    In this case the length is 2.6-1=1.6
    Then divided by the length of each one 1.6/.4=4
    Thus, a use of 4 intervals.

    By the trapezoid rule we need to find,
    f(1),f(1.4),f(1.8),f(2.2),f(2.6)

    We note that,
    f(1)=(1)^2*sin(1)=0.84
    f(1.4)=(1.4)^2*sin(1.4)=1.93
    f(1.8)=(1.8)^2*sin(1.8)=3.16
    f(2.2)=(2.2)^2*sin(2.2)=3.91
    f(2.6)=(2.6)^2*sin(2.6)=3.48

    Now we double the inner terms (look at formula)
    Thus,
    f(1)=.84
    2f(1.4)=3.86
    2f(1.8)=7.32
    2f(2.2)=7.82
    f(2.6)=3.48

    Add them together,
    Result=23.32

    Now multiply it by the length of each interval, (.4)
    Thus,
    9.33
    Finally multiply by 1/2,
    4.67
    That is the approximate answer.
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  3. #3
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    Thanks for that quick response! How come I got this answer though:
    Attached Thumbnails Attached Thumbnails Integrate functions and use calculus to solve problems-integrate-functions.jpg  
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  4. #4
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    Hello, norivea!

    Use a numerical method to approximate the area enclosed by:
    the graph of y = x▓ sin x, the x axis and the lines x = 1 and x = 2.6
    Use an interval of 0.4

    There are many available methods.

    I'll use left-endpoints.

    . f(1) . = . .1▓Ěsin(1) . . .0.8415
    f(1.4) .= .1.4▓Ěsin(1.4) . .1.9315
    f(1.8) .= .1.8▓Ěsin(1.8) . .3.1553
    f(2.2) .= .2.2▓Ěsin(2.2) . .3.9131
    f(2.4) .= .2.4▓Ěsin(2.4) . .3.8907
    f(2.6) .= .2.6▓Ěsin(2.6) . .3.4845

    Area . .(0.8415 + 1.9315 + 3.1553 + 3.9131 + 3.8907 + 3.4845) x 0.4 .= .6.88676

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  5. #5
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    Quote Originally Posted by norivea View Post
    Thanks for that quick response! How come I got this answer though:
    Your answer is 4.4646 mine is 4.67
    That is close enough!
    You did good.
    ---
    Note Soroban used Rectangular rule which is completelty innacturate unless you are using many. Your and mine method using trapezoids is much more accurate.

    The actual value is 4.62
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  6. #6
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    Quote Originally Posted by norivea View Post
    Thanks for that quick response! How come I got this answer though:
    Because ImperfectHacker evaluated f(1.8) incorrectly. Your solution is correct.

    RonL
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, norivea!


    There are many available methods.

    I'll use left-endpoints.

    . f(1) . = . .1²·sin(1) . . .0.8415
    f(1.4) .= .1.4²·sin(1.4) . .1.9315
    f(1.8) .= .1.8²·sin(1.8) . .3.1553
    f(2.2) .= .2.2²·sin(2.2) . .3.9131
    f(2.4) .= .2.4²·sin(2.4) . .3.8907
    f(2.6) .= .2.6²·sin(2.6) . .3.4845

    Area . .(0.8415 + 1.9315 + 3.1553 + 3.9131 + 3.8907 + 3.4845) x 0.4 .= .6.88676
    When using the left end point method we dont use your last point f(2.6),
    also all the increments should be equal.

    So your solution should look like:


    . f(1) . = . .1²·sin(1) . . .0.8415
    f(1.4) .= .1.4²·sin(1.4) . .1.9315
    f(1.8) .= .1.8²·sin(1.8) . .3.1553
    f(2.2) .= .2.2²·sin(2.2) . .3.9131

    Area . .(0.8415 + 1.9315 + 3.1553 + 3.9131) x 0.4 .=3.94

    RonL
    Last edited by CaptainBlack; October 10th 2006 at 02:47 AM.
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  8. #8
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    Software gives answer of 4.53
    Which matches my answer.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Software gives answer of 4.53
    Which matches my answer.
    The actual value of the integral is of no relevance here, we have an
    excercise in numerical integration with a specified step size. The
    answer for any particular method is very unlikely to agree with the
    answer obtained analyticaly.

    Also 4.46 is closer to 4.53 than your answer of 4.67.

    RonL
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  10. #10
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    I now realized what the user was complaining about.
    When it wrote,
    4.464646
    I saw that as,
    4.646464
    And that was close to my answer = 4.67
    But that is not.
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