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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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May I ask what method of integration you chose?
first i used trig. sub. a-x^2, gives x=4sintheta, and then i used trig. integral of u-sub.
you could have done this with just an algebraic substitution

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Other answers:

\[u=16-x^2 => du=-2x dx \] \[x^2=16-u \] So \[\int\limits_{}^{}x^2 \sqrt{16-x^2} dx=\frac{-1}{2}\int\limits_{}^{}x^2 \sqrt{16-x^2} (-2x) dx\] \[=\frac{-1}{2}\int\limits_{}^{}(16-u) \sqrt{u} du \] \[=\frac{-1}{2}\int\limits_{}^{}(16 u^\frac{1}{2}-u^\frac{3}{2}) du\]
but it's x^3, not x^2
integral x^3*sqrt(16-x^2)dx
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
expression* that follows that first equal sign
do you see?
one more question
1 Attachment
Did you find P(x) as a 2nd degree taylor polynomial yet?
i do not know where to start......
\[P(x) \approx P(r)+P'(r)(x-r)+\frac{1}{2}P''(r)(x-r)^2 \] This is what they want you to use.
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
so for 5, i got p5(x)=120000-40000(x-5)+22500(x-5)^2, is this right so far?
\[P(x)=45000+\frac{45000}{120000}(x-5)+\frac{1}{2} \frac{45000}{-40000}(x-5)^2 \]
We are given \[P(5)=45000; \text{ and } 120000 \cdot P'(5)=45000 ; \text{ and } -40000 \cdot P''(5)=45000\]
That is how I got what was P'(5) and P''(5) and P(5)
We will use that P I wrote to approximate the value of portfolio bonds when r is 5.5.
oh are those commas?
I can't read.
ok what you wrote is good then
So use your equation not mine to approximate P(5.5)
but how will it turn the equation into just 1 number?solve for x?
i don't understand
\[P(x) \approx 120000-40000(x-5)+45000(x-5)^2 \] You can find P(5.5) It will result in one number
there is only one variable
and you are asked to replace that variable x with 5.5
this will give you P(5.5)
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
oh thank you so much, i thought i have to replace 5.5 in as in term of r
and don't forget the 1/2 part on that one part
but now i got it, 105625 would be the answer
yeah but you aren't given P''(5.5) or P'(5.5) or P(5.5) so that would be impossible
no that isn't the answer
don't forget the half part on that last term
\[P(x) \approx 120000-40000(x-5)+\frac{1}{2} 45000(x-5)^2 \]
which you wrote earlier has 22500 which is fine
I'm talking about for that 1/2*45000 part you simplified it earlier to 22500

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