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Math Help - How to find the Area between two curves...

  1. #1
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    How to find the Area between two curves...

    Hi guys,

    i'm currently learning substitution in definite integrals and one part of this chapter is finding the area between two curves.. however for this one problem i keep getting 0 as the area which i am not sure if it is right or not

    heres the problem

    basically, find the areas of the shaded region
    (i can't draw it so i'll just inform you of the upper and lower curves)

    f(x) = y=1
    g(x) = y = (cosx)^2

    First step i did was i found the limits of integration. To do this I solved y = 1 and y = (cosx)^2 simultaneously for x getting...

    (cosx)^2 = 1
    Hence, x = 0,pie making a = 0 and b = pie

    then I integrate...

    { (b,a) (f(x)-g(x))dx

    =

    { (pie,0) (1- (cosx)^2)dx
    or
    { (pie,0) (sinx)^2 dx


    u = sinx du = dx

    { 0,0 u^2 du

    which EQUALS 0....

    Is that right? I dont think the AREA is 0 though....what's wrong?

    Thanks in advance




    EDIT**** FOUND OUT THE PROBLEM FROM BELOW, BUT I STILL DONT GEt IT ~ ANSWER MY 2nD QUESTION WHEN I POSTED AGAIN!!!
    Last edited by pghaffari; February 16th 2009 at 09:04 AM.
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  2. #2
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    Your mistake is here:
    u = sinx du = dx

    It should be this way:
    u = sin(x) => du = cos(x) dx holds.

    But you should do more that above substitution. Use some trigonmetric relations to introduce a cos(x) in the integral (for example sin(2x)=2sin(x)cos(x))
    I am just feeling that then you can use integration by parts to get the answer, which is:

    1/2 (x - sin(x)cos(x))

    Good luck.

    -O
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  3. #3
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    wow see i fail at substitution, i was absent the day my teacher taught it

    can you go over it for me briefly?

    say i have u = sinx
    when you find du, does that mean find the derivative of u? aka sinx --> -cosx ?


    Also, work the problem out for me after that...I still dont get it :\ THX
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  4. #4
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    Reply

    Quote Originally Posted by pghaffari View Post
    Hi guys,

    i'm currently learning substitution in definite integrals and one part of this chapter is finding the area between two curves.. however for this one problem i keep getting 0 as the area which i am not sure if it is right or not

    heres the problem

    basically, find the areas of the shaded region
    (i can't draw it so i'll just inform you of the upper and lower curves)

    f(x) = y=1
    g(x) = y = (cosx)^2

    First step i did was i found the limits of integration. To do this I solved y = 1 and y = (cosx)^2 simultaneously for x getting...

    (cosx)^2 = 1
    Hence, x = 0,pie making a = 0 and b = pie

    then I integrate...

    { (b,a) (f(x)-g(x))dx

    =

    { (pie,0) (1- (cosx)^2)dx
    or
    { (pie,0) (sinx)^2 dx


    u = sinx, du = dx (HOW?)

    { 0,0 u^2 du

    which EQUALS 0....

    Is that right? I dont think the AREA is 0 though....what's wrong?

    Thanks in advance
     \int\limits_0^\pi  {\left( {1 - \cos ^2 x} \right)} dx \hfill \\

     = \int\limits_0^\pi  {\left( {\sin ^2 x} \right)} dx \hfill \\

    = \frac{1}<br />
{2}\int\limits_0^\pi  {\left( {1 - \cos 2x} \right)} dx{\text{  }} \hfill \\

    \left[ {{\text{because,  }}\cos 2x = 1 - 2\sin ^2 x \Rightarrow \sin ^2 x = \frac{{1 - \cos 2x}}<br />
{2}} \right] \hfill \\

    = \frac{1}<br />
{2}\left[ {x - 2\sin 2x} \right]_0^\pi   \hfill \\

     = \frac{1}<br />
{2}\left[ {\pi  - 0} \right] = \frac{\pi }<br />
{2} \hfill \\ <br />

    Please see attached graph.
    Attached Thumbnails Attached Thumbnails How to find the Area between two curves...-graph26.jpg  
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  5. #5
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    What if i were to use SUBSTITUTION? How would I do it?
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  6. #6
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    Actually in Shyam's solution there is a little substitution. That is:
    Let u=2x, then du=2dx, and it goes from there.

    In general:
    u=f(x), then du=f'(x)dx
    Example:
    int{ (2x+1)*sin(x^2+x) dx } = int{ sin(u) du }
    Because:
    Let u=x^2+x, then du=(2x+1) dx, which in the integral.

    -O
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  7. #7
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    Ok what about this one

    { (2,-2) (4x^2 - x^4) dx

    u = 4x^2 - x^4 and du = (8x-4x^3)dx

    Yes/no?

    If so, what do i do after I substitute it back in..as in..how do i solve that question ~~~

    TY
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  8. #8
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    Quote Originally Posted by pghaffari View Post
    Ok what about this one

    { (2,-2) (4x^2 - x^4) dx

    u = 4x^2 - x^4 and du = (8x-4x^3)dx

    Yes/no?

    If so, what do i do after I substitute it back in..as in..how do i solve that question ~~~

    TY
    No substitution here. see,

    \int\limits_{ - 2}^2 {\left( {4x^2  - x^4 } \right)} dx \hfill \\

    =\left[ {\frac{{4x^3 }}<br />
{3} - \frac{{x^5 }}<br />
{5}} \right]_{ - 2}^2

     {\text{Now,  FINISH  it}}{\text{.}} \hfill \\
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  9. #9
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    You don't use subs. for all int questions. The last one u sent needs no subs. It's easy.

    { (2,-2) (4x^2 - x^4) dx
    = (4/3)x^3 - (1/5)x^5 THAT'S IT.

    Do not make easy problems unnecessarily difficult.
    I would suggest you find and study the class notes you've missed on integrals. Cause your asking very basic questions.

    All the best..

    -O
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  10. #10
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    ahhhhh, sorry i didnt see how easy it is...

    i was absent when we learned this in class sorry i see how stupid my questions are now

    ty <3


    okk im getting the hang of it now, i did the next problem by myself and got it right

    TY
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  11. #11
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    yay im getting the rest right :P

    i find it funny how once you understand that basic principle you can answer any problem of its sort ;]
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