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## anonymous 2 years 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 Delete Cancel Submit

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1. myininaya
• 2 years ago
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May I ask what method of integration you chose?

2. anonymous
• 2 years 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
• 2 years ago
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you could have done this with just an algebraic substitution

4. myininaya
• 2 years 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. anonymous
• 2 years ago
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but it's x^3, not x^2

6. anonymous
• 2 years ago
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integral x^3*sqrt(16-x^2)dx

7. myininaya
• 2 years 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
• 2 years ago
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expression* that follows that first equal sign

9. myininaya
• 2 years ago
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do you see?

10. anonymous
• 2 years ago
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one more question

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

12. anonymous
• 2 years ago
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i do not know where to start......

13. myininaya
• 2 years 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
• 2 years 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. anonymous
• 2 years 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
• 2 years ago
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$P(x)=45000+\frac{45000}{120000}(x-5)+\frac{1}{2} \frac{45000}{-40000}(x-5)^2$

17. myininaya
• 2 years 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
• 2 years ago
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That is how I got what was P'(5) and P''(5) and P(5)

19. myininaya
• 2 years 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
• 2 years ago
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oh are those commas?

21. myininaya
• 2 years ago
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lol

22. myininaya
• 2 years ago
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23. myininaya
• 2 years ago
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ok what you wrote is good then

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

25. anonymous
• 2 years ago
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but how will it turn the equation into just 1 number?solve for x?

26. anonymous
• 2 years ago
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i don't understand

27. myininaya
• 2 years 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
• 2 years ago
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there is only one variable

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

30. myininaya
• 2 years ago
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this will give you P(5.5)

31. myininaya
• 2 years 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. anonymous
• 2 years 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
• 2 years ago
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and don't forget the 1/2 part on that one part

34. anonymous
• 2 years ago
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but now i got it, 105625 would be the answer

35. myininaya
• 2 years 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
• 2 years ago
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37. myininaya
• 2 years ago
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don't forget the half part on that last term

38. myininaya
• 2 years ago
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$P(x) \approx 120000-40000(x-5)+\frac{1}{2} 45000(x-5)^2$

39. myininaya
• 2 years ago
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which you wrote earlier has 22500 which is fine

40. myininaya
• 2 years 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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