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anonymous

  • one year ago

Evaluate the definite integrals. \[\int\limits_{0}^{4} (\frac{ 3 }{ 2x+1 })dx\]

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  1. anonymous
    • one year ago
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    u=2x+1 du=2dx ??

  2. ganeshie8
    • one year ago
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    thats a good start

  3. anonymous
    • one year ago
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    New limit x=0, u=1 x=4, u=9

  4. anonymous
    • one year ago
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    So it would like this?.. \[\int\limits_{1}^{9}\frac{ 3 }{ u }du\] ??

  5. ganeshie8
    • one year ago
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    not quite, there is a mistake with differential du=2dx dx = ?

  6. anonymous
    • one year ago
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    dx= 1/2 du ?

  7. ganeshie8
    • one year ago
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    Yes

  8. anonymous
    • one year ago
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    Oh so it would be: \[\int\limits_{1}^{9}\frac{ 3 }{ 2u }du\]

  9. anonymous
    • one year ago
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    Then I got 3/2ln9 as the final answer..?

  10. ganeshie8
    • one year ago
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    looks good!

  11. anonymous
    • one year ago
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    Is it possible to reduce that even more?

  12. ganeshie8
    • one year ago
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    use the log property \[\large n\,\log x ~~=~~\log x^n \]

  13. anonymous
    • one year ago
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    I'm not sure how to use it correctly...

  14. ganeshie8
    • one year ago
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    \[\large \frac{3}{2}\,\ln(9) ~~=~~\ln (9^{\frac{3}{2}}) \]

  15. anonymous
    • one year ago
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    Then what happens?...

  16. ganeshie8
    • one year ago
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    \[\large \frac{3}{2}\,\ln(9) ~~=~~\ln (9^{\frac{3}{2}}) =\ln({\sqrt{9~}}^3)\]

  17. anonymous
    • one year ago
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    Oh I see! It would be 3ln3!

  18. ganeshie8
    • one year ago
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    Yep!

  19. anonymous
    • one year ago
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    Thank you! :)

  20. ganeshie8
    • one year ago
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    yw!

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