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BaoZhu

  • one year ago

integral x^3(sqrt(16-x^2)dx can someone check if my answer is right, (-1024/3)((sqrt(16-x^2)/(3)))^3+(1024/5)((sqrt(16-x^2)/(3)))^5+C

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  1. myininaya
    • one year ago
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    May I ask what method of integration you chose?

  2. BaoZhu
    • one year ago
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    first i used trig. sub. a-x^2, gives x=4sintheta, and then i used trig. integral of u-sub.

  3. myininaya
    • one year ago
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    you could have done this with just an algebraic substitution

  4. myininaya
    • one year ago
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    \[u=16-x^2 => du=-2x dx \] \[x^2=16-u \] So \[\int\limits_{}^{}x^2 \sqrt{16-x^2} dx=\frac{-1}{2}\int\limits_{}^{}x^2 \sqrt{16-x^2} (-2x) dx\] \[=\frac{-1}{2}\int\limits_{}^{}(16-u) \sqrt{u} du \] \[=\frac{-1}{2}\int\limits_{}^{}(16 u^\frac{1}{2}-u^\frac{3}{2}) du\]

  5. BaoZhu
    • one year ago
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    but it's x^3, not x^2

  6. BaoZhu
    • one year ago
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    integral x^3*sqrt(16-x^2)dx

  7. myininaya
    • one year ago
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    I missed an x in that first part but and that next equation you see i have x^2 times x which is x^3

  8. myininaya
    • one year ago
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    expression* that follows that first equal sign

  9. myininaya
    • one year ago
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    do you see?

  10. BaoZhu
    • one year ago
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    one more question

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  11. myininaya
    • one year ago
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    Did you find P(x) as a 2nd degree taylor polynomial yet?

  12. BaoZhu
    • one year ago
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    i do not know where to start......

  13. myininaya
    • one year ago
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    \[P(x) \approx P(r)+P'(r)(x-r)+\frac{1}{2}P''(r)(x-r)^2 \] This is what they want you to use.

  14. myininaya
    • one year ago
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    They want you to use r as 5 and plug in the values they gave you for P'(5) and P''(5) Then they want you to find P(5.5) approximately using the resulting equation

  15. BaoZhu
    • one year ago
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    so for 5, i got p5(x)=120000-40000(x-5)+22500(x-5)^2, is this right so far?

  16. myininaya
    • one year ago
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    \[P(x)=45000+\frac{45000}{120000}(x-5)+\frac{1}{2} \frac{45000}{-40000}(x-5)^2 \]

  17. myininaya
    • one year ago
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    We are given \[P(5)=45000; \text{ and } 120000 \cdot P'(5)=45000 ; \text{ and } -40000 \cdot P''(5)=45000\]

  18. myininaya
    • one year ago
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    That is how I got what was P'(5) and P''(5) and P(5)

  19. myininaya
    • one year ago
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    We will use that P I wrote to approximate the value of portfolio bonds when r is 5.5.

  20. myininaya
    • one year ago
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    oh are those commas?

  21. myininaya
    • one year ago
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    lol

  22. myininaya
    • one year ago
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    I can't read.

  23. myininaya
    • one year ago
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    ok what you wrote is good then

  24. myininaya
    • one year ago
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    So use your equation not mine to approximate P(5.5)

  25. BaoZhu
    • one year ago
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    but how will it turn the equation into just 1 number?solve for x?

  26. BaoZhu
    • one year ago
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    i don't understand

  27. myininaya
    • one year ago
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    \[P(x) \approx 120000-40000(x-5)+45000(x-5)^2 \] You can find P(5.5) It will result in one number

  28. myininaya
    • one year ago
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    there is only one variable

  29. myininaya
    • one year ago
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    and you are asked to replace that variable x with 5.5

  30. myininaya
    • one year ago
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    this will give you P(5.5)

  31. myininaya
    • one year ago
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    It is just like if i asked you to evaluate f(2) given f(x)=x-5 you would say f(2)=2-5=-3

  32. BaoZhu
    • one year ago
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    oh thank you so much, i thought i have to replace 5.5 in as in term of r

  33. myininaya
    • one year ago
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    and don't forget the 1/2 part on that one part

  34. BaoZhu
    • one year ago
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    but now i got it, 105625 would be the answer

  35. myininaya
    • one year ago
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    yeah but you aren't given P''(5.5) or P'(5.5) or P(5.5) so that would be impossible

  36. myininaya
    • one year ago
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    no that isn't the answer

  37. myininaya
    • one year ago
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    don't forget the half part on that last term

  38. myininaya
    • one year ago
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    \[P(x) \approx 120000-40000(x-5)+\frac{1}{2} 45000(x-5)^2 \]

  39. myininaya
    • one year ago
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    which you wrote earlier has 22500 which is fine

  40. myininaya
    • one year ago
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    I'm talking about for that 1/2*45000 part you simplified it earlier to 22500

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