Question

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

Answer 1

May I ask what method of integration you chose?

Answer 2

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

Answer 3

you could have done this with just an algebraic substitution

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

expression* that follows that first equal sign

Answer 9

do you see?

Answer 10

one more question

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

oh are those commas?

Answer 21

lol

Answer 22

I can't read.

Answer 23

ok what you wrote is good then

Answer 24

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

Answer 25

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

Answer 26

i don't understand

Answer 27

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

Answer 28

there is only one variable

Answer 29

and you are asked to replace that variable x with 5.5

Answer 30

this will give you P(5.5)

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

but now i got it, 105625 would be the answer

Answer 35

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

Answer 36

no that isn't the answer

Answer 37

don't forget the half part on that last term

Answer 38

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

Answer 39

which you wrote earlier has 22500 which is fine

Answer 40

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