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anonymous

  • 5 years ago

can someone show me how to find the indefinite integral of 1-y^2/3y

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  1. anonymous
    • 5 years ago
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    do you mean (1-y^2)/3y or 1- (y^2/3y)?

  2. anonymous
    • 5 years ago
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    first one

  3. anonymous
    • 5 years ago
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    ok.... first factor out your constant: (1/3) integral (1-y^2)/y dy expand the fraction: (1/3) integral (1/y) - y dy now integrate term by term: (1/3) integral (1/y) dy -(1/3) integral y dy integral of 1/y is ln(y) and integral of y is ((y^2)/2) solution: (1/3)ln(y) - (y^2)/6 + C

  4. anonymous
    • 5 years ago
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    does that make sense?

  5. anonymous
    • 5 years ago
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    how does 1/3 y dy become y^2/2

  6. anonymous
    • 5 years ago
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    I factored out the 1/3 first because it's a constant, then integrated y using the power rule, so (Y^2)/2, then multiply by (1/3) which gives (y^2)/6 that you see in the solution.

  7. anonymous
    • 5 years ago
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    Yes! You are great! It made perfect sense, thanks a bunch

  8. anonymous
    • 5 years ago
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    one more question: Find antiderivative of each derivative that satisfies the given condition: dR/dx = (1-x^4)/x^3

  9. anonymous
    • 5 years ago
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    oops left out R(1) = 4

  10. anonymous
    • 5 years ago
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    expand fraction in integrand and seperate: integral 1/x^3 dx - integral x dx evaluate using power rule on each: -1/(2x^2) - ((x^2)/2) + C evaluate using given initial condition of R(1) = 4 and solving for C: R(x) = -1/(2x^2) - ((x^2)/2) + 5

  11. anonymous
    • 5 years ago
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    thanks i understand, but can I express -1/2x^-2 instead of factoring out the negative?

  12. anonymous
    • 5 years ago
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    oops.. my bad.. change 2x to be in denominator

  13. anonymous
    • 5 years ago
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    i got it now

  14. anonymous
    • 5 years ago
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    ok, good ; )

  15. anonymous
    • 5 years ago
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    thanks a lot... been great help.. is this open 24/7? I might need to come back again later today

  16. anonymous
    • 5 years ago
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    i think so, depends on who or how many people are online here i guess...

  17. anonymous
    • 5 years ago
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    ok

  18. anonymous
    • 5 years ago
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    \[\int\limits_{4}^{1} (5x+3) dx \] evaluate the integral... I am back already :)

  19. anonymous
    • 5 years ago
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    oops the other way around between 1 and 4

  20. anonymous
    • 5 years ago
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    ok, so...... integral of 5x + 3 = (5/2)x^2 + 3x...now... ((5/2)(4^2) + 3(4)) - ((5/2)(1^2) + 3(1)) = 93/2

  21. anonymous
    • 5 years ago
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    thats right.. I left out the 3x.. thanks!

  22. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{1} e^x dx\]

  23. anonymous
    • 5 years ago
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    is it e -1 =1.718

  24. anonymous
    • 5 years ago
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    correct.

  25. anonymous
    • 5 years ago
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    thanks.. one last question for you

  26. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{2} (2x^-2 -3)dx\]

  27. anonymous
    • 5 years ago
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    evaluate the integral

  28. anonymous
    • 5 years ago
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    is that supposed to be 2x^2 - 3?

  29. anonymous
    • 5 years ago
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    no its a negative exponent

  30. anonymous
    • 5 years ago
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    ok... so integral of 2x^(-2) - 3 = (-2/x) - 3x... then...((-2/2) - 3(2)) - ((-2/1) - 3(1)) = -7 - (-5) = -2

  31. anonymous
    • 5 years ago
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    this is where I get confused. since 2/x don't you apply ln|x| to 1/x?

  32. anonymous
    • 5 years ago
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    no, the result of integrating 2x^(-2) IS -2/x, you don't want to integrate it again, just evaluate it...

  33. anonymous
    • 5 years ago
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    oh i see...

  34. anonymous
    • 5 years ago
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    that answered my question..

  35. anonymous
    • 5 years ago
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    great.. have a great afternoon and thanks again

  36. anonymous
    • 5 years ago
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    1/ \[\sqrt[3]{t}\]

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