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Thread: 4 Integral Questions

  1. #1
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    4 Integral Questions

    few integration questions that are bugging me i have made attempt on all of them some of them are not even worth mentioning how i have tried it because it got me no where near the answer.

    1. i am not even sure how to begin this one.

    $\displaystyle
    =\int_3^7 \sqrt(16t^2 + 8t + 1)dt
    $

    answer = 84

    attempt:

    $\displaystyle =\int_3^7 (16t^2 + 8t + 1)^1/2 dt $
    $\displaystyle =\int_3^7 \frac {(16t^2 + 8t + 1)^3/2}{3/2} dt $
    $\displaystyle =\int_3^7 \frac {2}{3} (16t^2 + 8t + 1)^3/2 dt$

    or other path i have tried as well... but wasn't sure how to exactly fully proceed

    $\displaystyle u = 16t^2 + 8t + 1 $
    $\displaystyle du = 32t + 8 dt $

    so i have tried this then...

    $\displaystyle \frac {1}{32t} + \frac {1}{8} du = dt $

    2. this one i am not sure if i copied wrong or something but when i attempt it i get the answer of 1

    question:
    $\displaystyle
    =\int_0^1 \frac{dx}{\sqrt(3 - 2x)}
    $

    answer = 0.73

    attempt:

    $\displaystyle
    u = (3-2x)
    $

    $\displaystyle
    du = -2dx
    $

    $\displaystyle
    -\frac{1}{2}du = dx
    $

    $\displaystyle
    =-\frac{1}{2}\int_0^1 u^-1/2 du = -\frac{1}{2}\int_0^1 \frac {u^1/2}{1/2} = -\frac{1}{2} * \frac{2}{1}(3-2x)^1/2
    $

    $\displaystyle
    = -1 \sqrt (3-2x)
    $

    3. this one gives me some problems as well. question says
    find the equation that passes thru (1,0) is tangent to the line $\displaystyle 6x + y = 6 $ and has $\displaystyle y" = \frac {12}{x^3} $


    answer: $\displaystyle \frac {6 - 6x}{x} $

    4. last question says
    find first quadrant area of $\displaystyle x + y + y^2 = 2 $


    answer: $\displaystyle \frac {7}{6} $

    attempt:
    first rewrite the equation to be

    $\displaystyle x = y + y^2 - 2 $

    got the limits using table of values and got result of (2,1)

    $\displaystyle A = \int_1^2 x dy $

    $\displaystyle = \int_1^2 (y + y^2 - 2)dy $

    $\displaystyle = \int_1^2 (\frac{y^2}{2} + \frac {y^3}{3} - 2y) $

    and then i am not sure if i am on the right path because it does not work out to correct answer.
    Last edited by araym1; Dec 13th 2009 at 05:54 PM.
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  2. #2
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    On #1, you seem to have forgotten the Chain Rule. Try completing the square under the radical and give some thought to a trigonometric substitution.
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  3. #3
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    On #2, you seem to have forgotten to change your limits when you changed variables.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    [quote=araym1;424449]few integration questions that are bugging me i have made attempt on all of them some of them are not even worth mentioning how i have tried it because it got me no where near the answer.

    1. i am not even sure how to begin this one. Begin by completing the square.

    $\displaystyle
    =\int_3^7 \sqrt(16t^2 + 8t + 1)dt
    $

    answer = 84

    attempt:

    $\displaystyle =\int_3^7 (16t^2 + 8t + 1)^1/2 dt $
    $\displaystyle =\int_3^7 \frac {(16t^2 + 8t + 1)^3/2}{3/2} dt $
    $\displaystyle =\int_3^7 \frac {2}{3} (16t^2 + 8t + 1)^3/2 dt$

    2. this one i am not sure if i copied wrong or something but when i attempt it i get the answer of 1

    question:
    $\displaystyle
    =\int_0^1 \frac{dx}{\sqrt(3 - 2x)}
    $

    answer = 0.73

    attempt:

    $\displaystyle
    u = (3-2x)
    $

    $\displaystyle
    du = -2dx
    $

    $\displaystyle
    \frac{-1}{2}du = dx
    $

    $\displaystyle
    =-\frac{1}{2}\int_0^1 u^-1/2 du = -\frac{1}{2}\int_0^1 \frac {u^1/2}{1/2} = -\frac{1}{2} * \frac{2}{1}(3-2x)^1/2
    $

    $\displaystyle
    = -1 \sqrt (3-2x)
    $Looks good. Now finish.

    3. this one gives me some problems as well. question says
    find the equation that passes thru (1,0) is tangent to the line $\displaystyle 6x + y = 6 $ and has $\displaystyle y" = \frac {12}{x^3} $

    answer: $\displaystyle \frac {6 - 6x}{x} $

    4. last question says
    find first quadrant area of $\displaystyle x + y + y^2 = 2 $

    answer: $\displaystyle \frac {7}{6} $

    attempt:
    first rewrite the equation to be

    $\displaystyle x = y + y^2 - 2 $

    got the limits using table of values and got result of (2,1)

    $\displaystyle A = \int_1^2 x dy $

    $\displaystyle = \int_1^2 (y + y^2 - 2)dy $

    $\displaystyle = \int_1^2 (\frac{y^2}{2} + \frac {y^3}{3} - 2y) $

    and then i am not sure if i am on the right path because it does not work out to correct answer.What's the answer?[/quote]..
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  5. #5
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    On #4, how did you decide on the limits? Did you report the WHOLE problem statement?
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  6. #6
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    Quote Originally Posted by TKHunny View Post
    On #4, how did you decide on the limits? Did you report the WHOLE problem statement?
    using a table of values, set x=0 then y=0 and yes whole problem reported and the answer to the ? is

    $\displaystyle \frac {7}{6} $
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  7. #7
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    wow i can not believe i have never thought of completing the square, but that totally worked, it was the key that i was missing all along. thanks to both of you
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  8. #8
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    Quote Originally Posted by araym1 View Post
    using a table of values, set x=0 then y=0 and yes whole problem reported and the answer to the ? is

    $\displaystyle \frac {7}{6} $
    Still insufficient. Both sections in Quadrant I are unbounded. Quadrant III would be a different story.

    Ah! I see the problem. Your algebra is bad. It should be x = 2 - y - y^2. Give it another go.
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  9. #9
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    Quote Originally Posted by TKHunny View Post
    Still insufficient. Both sections in Quadrant I are unbounded. Quadrant III would be a different story.

    Ah! I see the problem. Your algebra is bad. It should be x = 2 - y - y^2. Give it another go.
    that also works and it changed my limits to 0,1
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