Results 1 to 9 of 9

Math Help - 4 Integral Questions

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    4

    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.

    <br />
=\int_3^7 \sqrt(16t^2 + 8t + 1)dt<br />

    answer = 84

    attempt:

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

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

    so i have tried this then...

     \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:
    <br />
=\int_0^1 \frac{dx}{\sqrt(3 - 2x)}<br />

    answer = 0.73

    attempt:

    <br />
u = (3-2x)<br />

    <br />
du = -2dx<br />

    <br />
-\frac{1}{2}du = dx<br />

    <br />
=-\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<br />

    <br />
= -1 \sqrt (3-2x)<br />

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


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

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


    answer:  \frac {7}{6}

    attempt:
    first rewrite the equation to be

     x = y + y^2 - 2

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

     A = \int_1^2 x dy

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

     = \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; December 13th 2009 at 05:54 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    On #2, you seem to have forgotten to change your limits when you changed variables.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    [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.

    <br />
=\int_3^7 \sqrt(16t^2 + 8t + 1)dt<br />

    answer = 84

    attempt:

     =\int_3^7 (16t^2 + 8t + 1)^1/2 dt
     =\int_3^7 \frac {(16t^2 + 8t + 1)^3/2}{3/2} dt
     =\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:
    <br />
=\int_0^1 \frac{dx}{\sqrt(3 - 2x)}<br />

    answer = 0.73

    attempt:

    <br />
u = (3-2x)<br />

    <br />
du = -2dx<br />

    <br />
\frac{-1}{2}du = dx<br />

    <br />
=-\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<br />

    <br />
= -1 \sqrt (3-2x)<br />
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  6x + y = 6 and has  y" = \frac {12}{x^3}

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

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

    answer:  \frac {7}{6}

    attempt:
    first rewrite the equation to be

     x = y + y^2 - 2

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

     A = \int_1^2 x dy

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

     = \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]..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    On #4, how did you decide on the limits? Did you report the WHOLE problem statement?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2009
    Posts
    4
    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

     \frac {7}{6}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2009
    Posts
    4
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    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

     \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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Dec 2009
    Posts
    4
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral questions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 20th 2009, 08:56 PM
  2. Integral Questions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2009, 12:05 PM
  3. Few more integral questions.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 26th 2007, 11:32 PM
  4. 3 Integral questions
    Posted in the Calculus Forum
    Replies: 23
    Last Post: April 2nd 2007, 10:59 AM
  5. 2 Integral questions
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 1st 2006, 01:07 PM

Search Tags


/mathhelpforum @mathhelpforum