Results 1 to 15 of 15

Math Help - area problem

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    8

    area problem

    Find the area between y=(3+x^2)e^2^e and the x-axis from x = 0to x = 1.

    I have no clue how to solve this. Please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    Find the area between y=(3+x^2)e^2^e and the x-axis from x = 0to x = 1.

    I have no clue how to solve this. Please help.
    This function is looking fairly peculiar.

    y = (3 + x^2)e^{2^e}

    y = (3 + x^2)(e^2)^e = (3 + x^2)e^{2e}

    are both possible interpretations, but should there be an "x" somewhere in the exponentials?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2007
    Posts
    8
    Thats the way the problem is written. The only hint I received was
    (Hint: Use the distributive property after setting up the first integral.)Does is make any sense to you because I am so lost. I also have another problem. Thanks for any help you can offer.

    Find the area between f(x)=(x-3)^2 and the x-axis over the indicated interval, [2, 4].


    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    Find the area between y=(3+x^2)e^2^e and the x-axis from x = 0to x = 1.

    I have no clue how to solve this. Please help.
    Quote Originally Posted by italygirl9 View Post
    Thats the way the problem is written. The only hint I received was
    (Hint: Use the distributive property after setting up the first integral.)Does is make any sense to you because I am so lost.
    No, it doesn't make any sense to me. Well, here's how to do what you wrote:

    I'm just going to set the e^2^e thing as a constant c. Whichever form it is supposed to be e^{2^e} or (e^2)^e, it is simply a constant. So:

    \int_0^1(3 + x^2)c dx = c\int_0^1(3 + x^2) dx

    = c \int_0^13 dx + c \int_0^1x^2 dx

    = 3c \int_0^1 dx + c \int_0^1 x^2 dx

    Can you do it from here?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    Find the area between f(x)=(x-3)^2 and the x-axis over the indicated interval, [2, 4].
    \int_2^4(x - 3)^2 dx

    There are two ways to do this, depending on how complicated a method you prefer.

    Method 1:
    \int_2^4(x - 3)^2 dx = \int_2^4(x^2 - 6x + 9)dx

    = \int_2^4x^2~dx - 6 \int_2^4x~dx + 9 \int_2^4 dx

    Method 2:
    Let y = x - 3 \implies dy = dx
    \int_2^4(x - 3)^2 dx = \int_{-1}^1y^2~dy
    (Note how the limits of the integration change.)

    Both of these methods, of course, give you the same answer. If you haven't done substitutions yet, ignore method 2 for now. You'll see it again in due time.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2007
    Posts
    8
    Can you explain why it was set up that way? My answer I come up with is 2/3 is that correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    Can you explain why it was set up that way? My answer I come up with is 2/3 is that correct?
    That's indeed the answer. But...

    How would you set it up differently? One definition of the Riemann integral is that it is the area between the x-axis and the curve over the given domain.

    Seeing as you are confused, but got the correct answer you must have done something differently. Could you tell me what you did?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2007
    Posts
    8
    I replaced x-3 with u to simplify it.
    du=1dx
    since du=1dx then dx=du/1
    dx-du
    used the anti-derivative
    then i replaced u=x-3 and solved for x

    Is this the way you did it?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    I replaced x-3 with u to simplify it.
    du=1dx
    since du=1dx then dx=du/1
    dx-du
    used the anti-derivative
    then i replaced u=x-3 and solved for x

    Is this the way you did it?
    Okay, so that's method 2 then. I just wouldn't bother to replace the u with x again.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2007
    Posts
    8
    What is the answer for the first problem I asked you about. I come up with 1, but i don't think thats right at all.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    What is the answer for the first problem I asked you about. I come up with 1, but i don't think thats right at all.
    Quote Originally Posted by topsquark View Post
    I'm just going to set the e^2^e thing as a constant c. Whichever form it is supposed to be e^{2^e} or (e^2)^e, it is simply a constant. So:

    \int_0^1(3 + x^2)c dx = c\int_0^1(3 + x^2) dx

    = c \int_0^13 dx + c \int_0^1x^2 dx

    = 3c \int_0^1 dx + c \int_0^1 x^2 dx
    = 3c x |_0^1 + c \cdot \frac{1}{3}x^3 |_0^1

    = 3c (1 - 0) + \frac{c}{3}(1^3 - 0^3)

    = 3c + \frac{c}{3} = \frac{10c}{3}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Aug 2007
    Posts
    8
    How do you get all the symbols in your replys?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    How do you get all the symbols in your replys?
    Move your mouse pointer over the symbols and left-click. You should get a box coming up with the code. Otherwise, look at the first few posts in this thread.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Aug 2007
    Posts
    8
    I tried with the symbols and for some reason it won't work. Can you put the solution for the 2/3 in the symbols on the computer. I have it on my scrap paper, but I would like to have the whole problem worked out on the computer? I really do thank you very much for you help with these problems. This is the last class I need to graduate with my 4 year degree. I have taken calculus twice now.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by italygirl9 View Post
    I tried with the symbols and for some reason it won't work. Can you put the solution for the 2/3 in the symbols on the computer. I have it on my scrap paper, but I would like to have the whole problem worked out on the computer? I really do thank you very much for you help with these problems. This is the last class I need to graduate with my 4 year degree. I have taken calculus twice now.
    If I am right, this is what you did:
    \int_2^4 (x - 3)^2 dx

    Let u = x - 3 \implies du = dx

    Thus
    \int (x - 3)^2 dx = \int u^2 du

    = \frac{1}{3}u^3 + C = \frac{1}{3}(x - 3)^3 + C

    So
    \int_2^4 (x - 3)^2 dx = \frac{1}{3}(4 - 3)^3 - \frac{1}{3}(2 - 3)^3

    = \frac{1}{3} - \frac{-1}{3} = \frac{2}{3}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area problem
    Posted in the Geometry Forum
    Replies: 7
    Last Post: August 5th 2011, 02:19 PM
  2. Differential eq problem and area problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 5th 2009, 04:46 PM
  3. Replies: 4
    Last Post: December 16th 2008, 10:45 PM
  4. Please help- area problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 29th 2008, 11:49 AM
  5. Area Problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 18th 2007, 08:12 PM

Search Tags


/mathhelpforum @mathhelpforum