anonymous
  • anonymous
Express the integrand as a sum of partial fractions and evaluate the integral. Integral from 0 to 1 of (x^3)/ (x^2+8x+16) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{0}^{1} \frac{ x^{3} }{ x ^{2}+8x+16}dx\] I've gotten as far as : \[\int\limits_{0}^{1} x+1+\frac{ 256 }{ (x+4)^{2} } +\frac{ 51 }{ (x+4)} dx\]
SolomonZelman
  • SolomonZelman
I think that is incorrect...
anonymous
  • anonymous
Ok, then how do I figure out A and B if the denominator is a perfect square?

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SolomonZelman
  • SolomonZelman
before I do division, this is how you do it one fraction is (x+4) and the other is (x+4)^2
SolomonZelman
  • SolomonZelman
(I am talking denominators)
anonymous
  • anonymous
Right
SolomonZelman
  • SolomonZelman
|dw:1450124479892:dw|
anonymous
  • anonymous
Oh my gosh, thank you. I totally messed up the long division.
SolomonZelman
  • SolomonZelman
So, the result for division part is?
anonymous
  • anonymous
x-8- ((8x+16)/(x^2+8x+16))
SolomonZelman
  • SolomonZelman
Hold, I think there is a problem...
SolomonZelman
  • SolomonZelman
my division was faulty... I will erase my mess and put up the actual thing
anonymous
  • anonymous
I'm not sure where the -48x+128 came from
anonymous
  • anonymous
*-128
SolomonZelman
  • SolomonZelman
|dw:1450125712980:dw|
SolomonZelman
  • SolomonZelman
I will repost the division later if you want (let me know), but the remainder I get 48x+128 and the quotient x-8
anonymous
  • anonymous
That's beautiful. So the remainder is +48x+128. Okey doke.
SolomonZelman
  • SolomonZelman
I need to devode more time to hand-work processes... I totally abndoned that lately:)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \int_0^1~x-8+\frac{48x+128} {x^2+8x+16}dx }\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \int_0^1~x-8~dx }\) you can do yourself
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \int_0^1~\frac{48x+128} {x^2+8x+16}dx }\) is what we need to solve
anonymous
  • anonymous
Right. ok, so then the denominator reduces to (x+4)^2 and when doing the partial fractions it looks like : \[\frac{ A }{ (x+4)^{2} } +\frac{ B }{ (x+4) }\]
SolomonZelman
  • SolomonZelman
yes totally.... (I would usually put (x+4) denominator before (x+4)^2, but that doesn';t matter
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{48x+128} {x^2+8x+16}=\frac{A}{(x+4)^2} +\frac{B}{x+4}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{48x+128} {x^2+8x+16}=\frac{A}{(x+4)^2} +\frac{B}{x+4}}\) multiply times \(x^2+8x+16\). \(\large\color{#000000 }{ \displaystyle 48x+128=A +(x+4)B}\)
SolomonZelman
  • SolomonZelman
now you can do the COEFFICIENTS way, or plug in different x-values.
anonymous
  • anonymous
Coefficients?
SolomonZelman
  • SolomonZelman
(I apologize for messing up with the division part)
SolomonZelman
  • SolomonZelman
yes, coefficients: \(\large\color{#000000 }{ \displaystyle 48x+128=A +(x+4)B}\) \(\large\color{#000000 }{ \displaystyle 48x+128=A +Bx+4B}\) \(\large\color{#000000 }{ \displaystyle \color{blue}{48x}+\color{green}{128}=\color{blue}{Bx}+\color{green}{A +4B}}\)
SolomonZelman
  • SolomonZelman
So, B=48 and 128=A+4B --> 128=A+4•48 A=-64
SolomonZelman
  • SolomonZelman
So how we compared?
SolomonZelman
  • SolomonZelman
This is going to be very useful, to put the coefficients of each \(x^n\) correspondingly just as I did (especially if you solve huge auxilary DE, if you plan to take DE)
SolomonZelman
  • SolomonZelman
Do you want an example of "coefficients" like this: after we do this problem / now / never
anonymous
  • anonymous
I can't, gotta go for a final. If you could just explain how we integrate the partial fractions part, I'd be good.
anonymous
  • anonymous
I think I understand the coefficients thing a little bit.
SolomonZelman
  • SolomonZelman
In any case, the equation that results when we solve for A and B is true \(\forall\)x, so you can plug in any x to solve it for A and B
SolomonZelman
  • SolomonZelman
So if coefficients are still aren't clear you can do the plugging
SolomonZelman
  • SolomonZelman
Do you see how we get the partial fractions? Then we have \(\large\color{#000000 }{ \displaystyle \int_0^1~x-8+\frac{48x+128} {x^2+8x+16}dx }\) \(\large\color{#000000 }{ \displaystyle \int_0^1~x-8+\frac{B} {x+4}+\frac {A}{(x+4)^2}~dx}\) \(\large\color{#000000 }{ \displaystyle \int_0^1~x-8+\frac{48} {x+4}+\frac {-64}{(x+4)^2}~dx}\)
SolomonZelman
  • SolomonZelman
can you integrate this from here?
SolomonZelman
  • SolomonZelman
(you may treat (x+4) as a single variable (because for u=x+4, du=dx), as regards to the last 2 terms)
anonymous
  • anonymous
I understand how to integrate definitely for the first part, but I'm not sure ... Ok, so the integration of the partial fractions would look like (48ln u -64lnu^2) from 0 to 1
SolomonZelman
  • SolomonZelman
-64 ln(u) ^2 ??!
SolomonZelman
  • SolomonZelman
What is the integral of x\(^{-2}\) ?
SolomonZelman
  • SolomonZelman
48ln(x+4) for the 3rd term, yes...
SolomonZelman
  • SolomonZelman
the last term integrated incorrectly
anonymous
  • anonymous
64u^-1
SolomonZelman
  • SolomonZelman
|dw:1450127025647:dw|
SolomonZelman
  • SolomonZelman
you are applying the power rule to the last term
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \int \frac{ -64 }{u^2} dx=64u^{-1} }\)
SolomonZelman
  • SolomonZelman
(I wrote without the +C)
anonymous
  • anonymous
Ok
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \left.\int_0^1~x-8+\frac{48} {x+4}+\frac {-64}{(x+4)^2}~dx \\ =\frac{x^2}{2}-8x+48\ln(x+4)+\frac{64}{x+4}\right|_{x=0}^{x=1} }\)
SolomonZelman
  • SolomonZelman
this is the correct result.
SolomonZelman
  • SolomonZelman
(Also, I am treating x+4, just like "u" or any other variable)
SolomonZelman
  • SolomonZelman
(be careful with treating a function as a varaible)
anonymous
  • anonymous
ok, then you get: -8[(1/2+(64/5)+48ln5)-{0+16+48ln4)]
anonymous
  • anonymous
crap. forgot the 8x
SolomonZelman
  • SolomonZelman
at x=1; (1/2)-8+48ln(5)+64/5 at x=0; 48ln(4)+16
anonymous
  • anonymous
answer = 48ln(5/4)-107/10
SolomonZelman
  • SolomonZelman
Yes, you have it correctly.
anonymous
  • anonymous
ok, thank you so much
SolomonZelman
  • SolomonZelman
yw
SolomonZelman
  • SolomonZelman
If I were to solve the following Differential Equation (DE): \(\large\color{#000000 }{ \displaystyle y''-3y' -4y=2x^3+x^2+12}\) \(y'' ~-~2{\rm nd~derivative~of~}y\) \(y' ~-~1{\rm st~derivative~of~}y\) then, the solution will be a combination (sum) of \(\large\color{#000000 }{ \displaystyle y_c(x)=c_1e^{r_1\cdot x}+ c_2e^{r_2\cdot x}}\), (where \(r_1\) and \(r_2\) are the roots of the auxilary equation \(r^2-3r-4\) (\(r^2\) for second derivative) (\(r^1\) for first derivative) (\(r^0\) (or no r) for just y (zeroth derivative)) AND \(\color{#000000 }{ \displaystyle y_g(x)}\) which (in this case) is in the form of \(\color{#000000 }{ Ax^3+Bx^2+Cx+D}\) (since the function on the right is 3rd degree polynomial) AND the solution to DE would be \(y(x)=y_c(x)+y_g(x)\). \(\color{#0000ff}{\Huge^{^\text{____________________________}}}\) So let's solve this DE: So, \(r^2-3r-4=0\) \(r_1=-1\quad{\rm and}\quad r_2=4\) \(y_c(x)=c_1e^{-x}+c_2e^{4x}\) \(\color{#ff0000}{\Huge{\text{_______________}}}\) The function \(y_g(x)\) is in the form of \(\color{#000000 }{ Ax^3+Bx^2+Cx+D}\). Therefore, We will find y", -3y' , and -4y of this, and solve for A,B,C and D. \(\color{#000000 }{ y\Longrightarrow \displaystyle Ax^3+Bx^2+Cx+D }\) therefore; \(\color{#000000 }{ -4y\Longrightarrow \displaystyle -4Ax^3-4Bx^2-4Cx-4D }\) \(\color{#000000 }{ y'\Longrightarrow \displaystyle 3Ax^2+2Bx+C }\) therefore; \(\color{#000000 }{ -3y'\Longrightarrow \displaystyle -9Ax^2-6Bx-3C }\) \(\color{#000000 }{ y''\Longrightarrow \displaystyle 6Ax+2B }\) So \(y''-3y'-4y\) becomes \(\small\color{#000000 }{ \displaystyle 6Ax+2B-9Ax^2-6Bx-3C -4Ax^3-4Bx^2-4Cx-4D}\) And we know that \(y''-3y'-4y=2x^3+x^2+12\) \(\color{#ff0000}{\LARGE {\text{___________________}}}\) \(\color{#ff0000}{\bf AND~~~HERE~~~IS~~~THE~~PART\\WITH~~~"COEFFICIENTS"}\) \(\color{#ff0000}{\Huge{^\text{___________________}}}\) Therefore, \(\color{#000000 }{ \displaystyle 6Ax+2B-9Ax^2-6Bx-3C -4Ax^3-4Bx^2-4Cx-4D \\ =2x^3+x^2+12}\) Now, let's group all terms by the powers of x; \(\color{#000000 }{ \displaystyle -4Ax^3-9Ax^2-4Bx^2+ 6Ax-6Bx-4Cx+2B-3C -4D \\ =2x^3+x^2+12}\) There is the \(x^1\) term on the right side... \(\color{#000000 }{ \displaystyle -4Ax^3-9Ax^2-4Bx^2+ 6Ax-6Bx-4Cx+2B-3C -4D \\ =2x^3+x^2+0x+ 12}\) \(\small \color{#000000 }{ \displaystyle \color{blue}{-4A}x^3+\color{green}{(-9A-4B)}x^2+ \color{red}{(6A-6B-4C)}x+\color{magenta}{2B-3C -4D} \\ =\color{blue}{2}x^3+\color{green}{1}x^2+\color{red}{0}x+\color{magenta}{ 12}}\) So we have: \(\color{#0000ff }{ \displaystyle -4A=2 }\) \(\color{green }{ \displaystyle -9A-4B=1 }\) \(\color{red}{ \displaystyle 6A-6B-4C=0 }\) \(\color{magenta }{ \displaystyle 2B-3C -4D=12 }\) (That just now was the essence of "coefficients") Then we can solve the system... \(\color{#0000ff }{ \displaystyle -4A=2~~~\longrightarrow A=-1/2 }\) \(\color{green }{ \displaystyle -9(-1/2)-4B=1~~~\longrightarrow B=7/8 }\) \(\color{red}{ \displaystyle 6(-1/2)-6(7/8)-4C=0~~~\longrightarrow C=-33/16 }\) \(\color{magenta }{ \displaystyle 2(7/8)-3(-33/16) -4D=12~~~\longrightarrow D=-65/64 }\) So, the \(y_g(x)=Ax^3+Bx^2+Cx+D\) function is, \(y_g(x)=\color{#0000ff }{\dfrac{-1}{2}}x^3+\color{green }{\dfrac{7}{8}}x^2+\color{red }{\dfrac{-33}{16}}x+\color{magenta }{\dfrac{-65}{64}}\) And therefore the answer (to DE) would be: \(y=y_g(x)+y_c(x)\) \(y=\left\{c_1e^{-x}+c_2e^{4x}\right\}+\left\{\color{#0000ff }{\dfrac{-1}{2}}x^3+\color{green }{\dfrac{7}{8}}x^2+\color{red }{\dfrac{-33}{16}}x+\color{magenta }{\dfrac{-65}{64}}\right\} \) (that is the genral solution to the DE and our DE was: \(\color{#000000 }{ \displaystyle y''-3y' -4y=2x^3+x^2+12}\) )
SolomonZelman
  • SolomonZelman
Now back to our division; \(\large\color{#000000 }{ \displaystyle x^3/(x^2+8x+16) }\)|dw:1450134662692:dw|

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