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sammii2u Group Title

evaluate the intergral -1 to 6 (x-2)/(x^2-5x-14) dx

  • one year ago
  • one year ago

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  1. sammii2u Group Title
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    \[\int\limits_{-1}^{6} \frac{ x-2 }{ x^2-5x-14 }\]

    • one year ago
  2. slaaibak Group Title
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    \[\int\limits_{-1}^{6} {x-2 \over (x+2)(x -7)} dx\] now do the whole partial fractions thing

    • one year ago
  3. sammii2u Group Title
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    \[\frac{ A }{ x-7 } + \frac{ B }{ x+2 }\] then getting everything with the same denominator I get.. \[x-2 = A(x+2) + B (x-7)\] then plugging in the different zeros, x=7 and x=-2 I solve for A and B. \[ A = \frac{ 5 }{ 9 }\] \[ B = \frac{ 4}{ 9 }\] Now the overall equation looks like: \[= \frac{ 5 }{ 9 } \int\limits_{}^{} \frac{ 1 }{ x-7 } dx + \frac{ 4 }{ 9 } \int\limits_{}^{} \frac{ 1 }{ x+2 } dx\] now what..? am I even doing this right?

    • one year ago
  4. Goten77 Group Title
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    hmm

    • one year ago
  5. Goten77 Group Title
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    yes... but in this form now its put into the ln form....

    • one year ago
  6. Goten77 Group Title
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    |dw:1358132021880:dw|

    • one year ago
  7. Goten77 Group Title
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    you got it into the like ... i cant think of the name but its a form name.... but you did it right

    • one year ago
  8. sammii2u Group Title
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    right. so it would be \[= \left[ \frac{ 5\ln |x-7| }{ 9 } + \frac{ 4\ln |x+2| }{ 9 } \right] _{-1} ^{6}\] and then I evaluate it at x=6 and then subtract x=-1 ?

    • one year ago
  9. slaaibak Group Title
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    yeah. nice job.

    • one year ago
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