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Math Help - Some integration questions

  1. #1
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    Some integration questions

    Just a few checks and a few questions..yes or no will do from most answers thanks =P

    1: d/dx of int between x and 1:t^2-2t+3

    I got x^2-2x+3..hard huh...

    2: find the area between x=3y and x=2+y^2

    I got 1/6

    3: find the volume of the region bounded by y=4-x^2 and y=0 which is rotated about the x axis:

    i got 6pi

    4: the base of a solid v is bounded by y=x^2 and y=2-x^2. find the volume if v has square cross sections

    I got 32/15 (2.1333333333)

    5: a particle has the velocity v(t)=t-1/4*t^2

    a) Find: the displacement between 0 and 6 seconds:
    b) Explain what this means
    c) find the distance travelled between 0 and 6 seconds
    d) find the average velocity between 0 and 4 seconds

    i got:
    a) 0
    b) it has arrived back at the starting point
    c) 36m
    d) help me with this one please

    6) use simpsons rule with 4 subsections to approximate the area under sqrt(x-1) between 2 and 5.
    I know the answer should be 14/3 however due to the errors in the small amount of subsections could my answer of 6.22 be counted as correct..please try this out yourself with only 4 subsections and see what you get.

    7) find the error bound in the result above

    i got 1.23596x10^-3, is this correct and can someone please provide the forumla for it too
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by deragon999 View Post
    Just a few checks and a few questions..yes or no will do from most answers thanks =P

    1: d/dx of int between x and 1:t^2-2t+3

    I got x^2-2x+3..hard huh...
    If "d/dx of int between x and 1:t^2-2t+3" means:

    \frac{d}{dx} \left( \int_x^1 t^2-2t+3 dt \right)

    then it may not be hard but your answer is wrong it's -x^2+2x-3.

    RonL
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  3. #3
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    CaptainB got the ball rolling. I'll do a couple more.

    Quote Originally Posted by deragon999 View Post
    [snip]
    2: find the area between x=3y and x=2+y^2

    I got 1/6 Mr F says: Correct. But you should say 1/6 square units.

    3: find the volume of the region bounded by y=4-x^2 and y=0 which is rotated about the x axis:

    i got 6pi Mr F says: Wrong. How did you get this value? {\color{red}V = \pi \int_{-2}^{2} y^2 \, dx = \pi \int_{-2}^{2} (4 - x^2)^2 \, dx = \, .....}

    4: the base of a solid v is bounded by y=x^2 and y=2-x^2. find the volume if v has square cross sections

    I got 32/15 (2.1333333333) Mr F says: Are you given a height? (The area of the base is 8/3).

    5: a particle has the velocity v(t)=t-1/4*t^2

    a) Find: the displacement between 0 and 6 seconds:
    b) Explain what this means
    c) find the distance travelled between 0 and 6 seconds
    d) find the average velocity between 0 and 4 seconds

    i got:
    a) 0 Mr F says: Correct. But you should say 0 m.
    b) it has arrived back at the starting point Mr F says: Correct.
    c) 36m Mr F says: Wrong. How did you get this value? Have you drawn a graph of x versus t? It starts at (0, 0), has a turning point at (4, 8/3) and a t-intercept at (6, 0) .....
    d) help me with this one please Mr F says: Average velocity = Displacement/Time = (8/3)/4 = .....

    [snip]
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  4. #4
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    Sorry question 1 is between 1 and x not between x and 1 so when you subtracted the two substitutions it all become negative.

    Question 2: yup ok i will =P(always do just writing quick to save time)

    Question 3: I accidentally said rotated about the x axis..its about the y axis lol..is 6pi correct now?

    Question 4: would it help if i mentioned that this is like a solid of revolution, but is not formed by revolving the graph(s) so the cross-section is not circle. Integrate the cross-section area with respect to x to get the volume.

    Does that help in your answer checking?

    Question 5: a) and b) thanks, c) yea i realised that today...first time my working was stupid..., d) thanks for that ill try it.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by deragon999 View Post
    Sorry question 1 is between 1 and x not between x and 1 so when you subtracted the two substitutions it all become negative.
    Are you supposed to do it that way? Are you not supposed to use the fundamental theorem of calculus:

    \frac{d}{dx} \left( \int_a^x f(t)\ dt \right) =f(x)

    RonL
    Last edited by CaptainBlack; August 6th 2008 at 09:08 PM.
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  6. #6
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    Quote Originally Posted by deragon999 View Post
    [snip]
    3: find the volume of the region bounded by y=4-x^2 and y=0 which is rotated about the x axis:



    i got 6pi
    [snip]
    Quote Originally Posted by deragon999 View Post
    [snip]
    Question 3: I accidentally said rotated about the x axis..its about the y axis lol..is 6pi correct now?

    [snip]
    Then V = \pi \int_{0}^{{\color{red}4}} x^2 \, dy = \pi \int_{0}^{{\color{red}4}} 4 - y \, dy . This is not equal to 6 \pi.
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  7. #7
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    Captainblack-yes you are correct, thats how i got my answer, i was just trying to explain why mr fantastic got negatives.

    Mr fantastic, yes..once again your correct..i did it between 2 and 0 as i imagined the line y=0 as x=0, which cut the rotated region in the wrong place.---so 8pi?

    Can you please also recheck the others, im sure im correct for question 4, if anyone knows how to find volumes of non rotated solids by integration feel free to try it out.
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