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Math Help - Integration

  1. #1
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    Integration

    1. Integrate following with respect to x:
    sech 2x

    2. Find the area of the finite region bounded by the curve with equation x^2 Ė y^2 = 4 and the line x = 5.
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  2. #2
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    \int{sech(2x)}dx

    Rewrite sech(2x)=\frac{2e^{2x}}{e^{4x}+1}

    Let u=e^{2x}, \;\ \frac{du}{2}=e^{2x}dx

    \int\frac{1}{u^{2}+1}du

    Now, see what the antiderivative is?.


    #2: Solve your equation for x in terms of y and integrate from 0 to 5.

    2\int_{0}^{5}\sqrt{4+y^{2}}dy
    Last edited by galactus; April 5th 2008 at 06:09 PM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    [tex]

    #2: Solve your equation for x in terms of y and integrate from 0 to 5.

    2\int_{0}^{5}\sqrt{4+y^{2}}dy
    There is a problem, I canít solve this.

    Ok, here is my work,

    Let, I = ∫(4+y^2)^1/2
    Let, x = (4+y^2)^3/2
    dx/dy = 3y (4+y^2)^1/2

    So: I = 1/3y (4+y^2)^3/2 + C

    With limits 0 Ė 5, I got the answer 10.41. But correct answer is 16.6.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geton View Post
    There is a problem, I canít solve this.

    Ok, here is my work,

    Let, I = ∫(4+y^2)^1/2
    Let, x = (4+y^2)^3/2
    dx/dy = 3y (4+y^2)^1/2

    So: I = 1/3y (4+y^2)^3/2 + C

    With limits 0 Ė 5, I got the answer 10.41. But correct answer is 16.6.
    where did you learn that integration technique?

    use a trig substitution, y = 2 \tan \theta
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    where did you learn that integration technique?
    I did a mistake.

    use a trig substitution, y = 2 \tan \theta
    By the way, trig substitution, can you explain for me? I donít get yet.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geton View Post
    By the way, trig substitution, can you explain for me? I donít get yet.
    do you mean you haven't done it in class, or you're just not good at it. in either case, look it up in your textbook. once you figure out what to substitute, it's pretty much the regular substitution technique. back-substituting can be somewhat annoying, but it is not that bad, and it can be avoided altogether with definite integrals as you see here.

    so let's start you off:

    we wish to evaluate  \int_0^5 \sqrt{4 + y^2}~dy

    Let y = 2 \tan \theta

    \Rightarrow dy = 2 \sec^2 \theta ~d \theta

    Change our limits to avoid back-substitution:

    when y = 0, \theta = \arctan \frac 02 = 0

    when y = 5, \theta = \arctan \frac 52

    So our integral becomes:

    4 \int_0^{\arctan 5/2} \sec^3 \theta ~d \theta


    Now continue
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