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Math Help - integration check 2

  1. #1
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    integration check 2

     <br />
\int_{-2}^{-1} \frac{1}{3-4x}<br />

     <br />
= (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2)))<br />

     <br />
= 0.049073661<br />

    correct? thank you
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  2. #2
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    Quote Originally Posted by jvignacio View Post
     <br />
\int_{-2}^{-1} \frac{1}{3-4x}<br />

     <br />
= (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2)))<br />

     <br />
= 0.049073661<br />

    correct? thank you
    No. The final numerical answer is wrong.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    No. The final numerical answer is wrong.
    the decimal places? is the substitution correct? where i enter -1 and -2 in for the x's?
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    the decimal places? is the substitution correct? where i enter -1 and -2 in for the x's?
    Quote Originally Posted by jvignacio View Post
     <br />
\int_{-2}^{-1} \frac{1}{3-4x}<br />

     <br />
= (\frac{-1}{4}log(3-4(-1))) - (\frac{-1}{4}log(3-4(-2)))<br />
Mr F says: Correct.


    = 0.049073661 Mr F says: Wrong.


    correct? thank you
    ..
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    ..
    wierd. my calculator is giving me that answer.
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    wierd. my calculator is giving me that answer.
    Then you're not entering the expression into your calculator correctly. Like the old saying says: Garbage in, garbage out.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Then you're not entering the expression into your calculator correctly. Like the old saying says: Garbage in, garbage out.
    hrmmm could be. ive tryed a few things and i keep getting the same answer.. ill keep trying
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  8. #8
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    The 'log' on your calculator means logarithm to base 10.
    Use the 'ln' button for e-based logarithm.
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  9. #9
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    Quote Originally Posted by qorilla View Post
    The 'log' on your calculator means logarithm to base 10.
    Use the 'ln' button for e-based logarithm.
    much appreciated! thank you
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