1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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