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

- anonymous

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- myininaya

May I ask what method of integration you chose?

- anonymous

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

- myininaya

you could have done this with just an algebraic substitution

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## More answers

- myininaya

\[u=16-x^2 => du=-2x dx \]
\[x^2=16-u \]
So \[\int\limits_{}^{}x^2 \sqrt{16-x^2} dx=\frac{-1}{2}\int\limits_{}^{}x^2 \sqrt{16-x^2} (-2x) dx\]
\[=\frac{-1}{2}\int\limits_{}^{}(16-u) \sqrt{u} du \]
\[=\frac{-1}{2}\int\limits_{}^{}(16 u^\frac{1}{2}-u^\frac{3}{2}) du\]

- anonymous

but it's x^3, not x^2

- anonymous

integral x^3*sqrt(16-x^2)dx

- myininaya

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

- myininaya

expression* that follows that first equal sign

- myininaya

do you see?

- anonymous

one more question

##### 1 Attachment

- myininaya

Did you find P(x) as a 2nd degree taylor polynomial yet?

- anonymous

i do not know where to start......

- myininaya

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

- myininaya

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

- anonymous

so for 5, i got p5(x)=120000-40000(x-5)+22500(x-5)^2, is this right so far?

- myininaya

\[P(x)=45000+\frac{45000}{120000}(x-5)+\frac{1}{2} \frac{45000}{-40000}(x-5)^2 \]

- myininaya

We are given \[P(5)=45000; \text{ and } 120000 \cdot P'(5)=45000 ; \text{ and } -40000 \cdot P''(5)=45000\]

- myininaya

That is how I got what was P'(5) and P''(5) and P(5)

- myininaya

We will use that P I wrote to approximate the value of portfolio bonds when r is 5.5.

- myininaya

oh are those commas?

- myininaya

lol

- myininaya

I can't read.

- myininaya

ok what you wrote is good then

- myininaya

So use your equation not mine to approximate P(5.5)

- anonymous

but how will it turn the equation into just 1 number?solve for x?

- anonymous

i don't understand

- myininaya

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

- myininaya

there is only one variable

- myininaya

and you are asked to replace that variable x with 5.5

- myininaya

this will give you P(5.5)

- myininaya

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

- anonymous

oh thank you so much, i thought i have to replace 5.5 in as in term of r

- myininaya

and don't forget the 1/2 part on that one part

- anonymous

but now i got it, 105625 would be the answer

- myininaya

yeah but you aren't given P''(5.5) or P'(5.5) or P(5.5) so that would be impossible

- myininaya

no that isn't the answer

- myininaya

don't forget the half part on that last term

- myininaya

\[P(x) \approx 120000-40000(x-5)+\frac{1}{2} 45000(x-5)^2 \]

- myininaya

which you wrote earlier has 22500 which is fine

- myininaya

I'm talking about for that 1/2*45000 part
you simplified it earlier to 22500

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