Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple definite integral

  1. Feb 26, 2009 #1


    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data
    Find the area bounded by the curve [tex]y=x^2-2[/tex], the y-axis, y=0 and y=1

    I got an answer, but when I checked it using graphmatica, it was wrong (also the result is logically too large). I cannot see where I went wrong though, so could someone please help me spot the mistake.

    2. Relevant equations

    [tex]\int_{a}^{b}f(y)dy = \left [ F(y) \right ]_{a}^{b}[/tex]

    [tex]\int_{a}^{b}(ax+b)^ndx=\left [ \frac{(ax+b)^{n+1}}{a(n+1)}\right ]_a^b[/tex]

    3. The attempt at a solution

    where [tex]x=\pm \sqrt{y+2}[/tex]

    [tex]A=\int_{0}^{1}\pm \sqrt{y+2}.dy[/tex]


    [tex]=2 \left[ \frac{(y+2)^{3/2}}{3/2} \right]_0^1[/tex]

    [tex]=2\left( \frac{3^{3/2}}{3/2} \right)[/tex]

  2. jcsd
  3. Feb 26, 2009 #2


    User Avatar
    Homework Helper

    hmm... i think it could be the factor of 2, the question says from y axis, so try only taking +sqrt
  4. Feb 26, 2009 #3


    User Avatar
    Homework Helper

    Uh no actually, there is something else going on here.

    This is why:

    Using graphmatica, I found [tex]\int_0^1 \sqrt{x+2}.dx \approx 1.5785[/tex]

    I changed the function to integrate because the program seems to only be able to integrate with respect to the x-axis. The function is still equivalent to the original question though (for positive values of y).

    So if I take the area of [tex]\int_0^1 \sqrt{y+2}.dy[/tex] in the positive quadrant only, then I get [tex]2\sqrt{3} \approx 3.464[/tex] which is far off the value given from graphmatica (and this answer doesn't seem correct either, from an elementary view of the approx area).

    The problem is somewhere else, and I can't seem to spot it...
  5. Feb 26, 2009 #4


    User Avatar
    Homework Helper

    Ahh I found where I went wrong:

    [tex]\left[ \frac{(y+2)^{3/2}}{3/2} \right]^1_0 \neq 2\sqrt{3}[/tex]

    I assumed substituting 0 into the primitive returned 0. I'll be sure not to make that mistake again..
  6. Feb 26, 2009 #5


    User Avatar
    Science Advisor

    That's makes no sense. You must integrate a function and "[itex]\pm\sqrt{y+2}[/itex]" is not a function. The x distance from one side of the graph to the other is [itex]2\sqrt{y+2}[/itex]. Further, [tex]\int_0^1 xdx[/itex] is NOT the same as [tex]\int_0^1\sqrt{y+ 2}dy[/tex]. If you let [tex]x= \sqrt{y+2}= (y+ 2)^{1/2}[/tex], then [tex]dx= (1/2)(y+2)^{-1/2}dy= 1/(2x)dy[/tex] so [tex]dy= 2xdx[/tex]. Also when y= 0, [itex]x= \sqrt{2}[/itex] and when y= 1, [itex]x= \sqrt{3}[/itex].
    [tex]\int_0^1 \sqrt{y+2}dy= 2\int_{\sqrt{2}}^{\sqrt{3}} x^2dx= \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right)[/tex] so your area is [itex](4/3)\left(\sqrt{8}- \sqrt{2}\right)[/itex].

    Last edited by a moderator: Feb 26, 2009
  7. Feb 26, 2009 #6


    User Avatar
    Homework Helper

    How so? if [itex]y=x^2-2[/itex] is a function, then if x is made the subject, why isn't it a function anymore?

    Yes I did this, but you could argue I made it in a less conventional (cheating) way :wink:

    The area you found was correct but you simplified it incorrectly.

    [tex]\frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right) \neq (4/3)\left(\sqrt{8}- \sqrt{2}\right)[/tex]

    The answer [itex]A=\frac{4}{3}(3\sqrt{3}-2\sqrt{2}) \approx 3.157[/itex] is also what Graphmatica gave me. Now the question as to whether the area is actually half of this or not. The question says it is also bounded by the y-axis, but [itex]y=x^2-2[/itex] has x values of both negative and positive between [itex]0<y<1[/itex] to the y-axis...
    Last edited: Feb 26, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook