Expert: Paul Klarreich - 5/23/2008

Question
QUESTION: Give that 7x-x^2 /(2-x) (x^2 +1) = A/(2-x) + Bx+C/(x^2 +1), determine the values of A,B and C. A curve has equation y = 7x-x^2 /(2-x) (x^2 +1). Prove that the are of the region bounded by the curve, the x-axis and the line x=1 is 7/2 ln2 - pi/4.
I am stuck with this, as i haven't used partial fraction with the Bx+C part for a while.
Thanks for your help!

ANSWER: Questioner:   jenny
Category:  Advanced Math
Private:  No
 
Subject:  Further Integration (i think)?
Question:  Give that 7x-x^2 /(2-x) (x^2 +1) = A/(2-x) + Bx+C/(x^2 +1), determine the values of A,B and C.

A curve has equation y = 7x-x^2 /(2-x) (x^2 +1). Prove that the are of the region bounded by the curve, the x-axis and the line x=1 is 7/2 ln2 - pi/4.
I am stuck with this, as i haven't used partial fraction with the Bx+C part for a while.
Thanks for your help!
.........................
Hi, Jenny,

This is going to be a problem, because the graph of

     7x-x^2
y = -------------
   (2-x)(x^2 +1)

has a vertical asymptote at x = 2.  You will have to be more specific about your boundaries if you want an area. [It would be nice if you proofread your question, too.]

As to the partial fractions, you can switch the signs on top and bottom and do this:

  x^2 - 7x         A      Bx + C
--------------- = ----- + ---------
(x - 2)(x^2 +1)   x - 2    x^2 + 1

  x^2 - 7x        Ax^2 + A + Bx^2 - 2Bx + Cx - 2C
--------------- = --------------------------------
(x - 2)(x^2 +1)       (x - 2)(x^2 + 1)

Equate coefficients:

Quad:  1 = A + B
Linear:  -7 = -2B + C
Const:   0 = A - 2C

A = 2C
C = 2B - 7

1 = A + B
1 = 2C + B

1 = 2(2B - 7) + B
1 = 4B - 14 + B
15 = 5B

B = 3
C = - 1
A = - 2


  x^2 - 7x        -2       3x - 1
--------------- = ----- + ---------
(x - 2)(x^2 +1)   x - 2    x^2 + 1


  x^2 - 7x        -2       3x          1
--------------- = ----- + --------- - -------
(x - 2)(x^2 +1)   x - 2    x^2 + 1    x^2 + 1

You split the last terms because you integrate them separately.  The three terms integrate as:

A log of (x - 2)
A log of (x^2 + 1)
An arctan(x)

That will have to be it for me -- I don't know what your left-and right-hand boundaries are.


---------- FOLLOW-UP ----------

QUESTION: Hi thank youu, what do u mean by left and right hand boundaries? Is it the limits at which you integrate? If that is so, I dont kno either because it wasnt provided in the question. I have attached the question as an image, if it helps. thank you again.

Answer
QUESTION: Hi thank youu, what do u mean by left and right hand boundaries? Is it the limits at which you integrate? If that is so, I dont kno either because it wasnt provided in the question. I have attached the question as an image, if it helps. thank you again.

A curve has equation y = (7x-x^2) /((2-x) (x^2 +1)). Prove that the area of the region bounded by the curve, the x-axis and the line x=1 is 7/2 ln2 - pi/4.


  x^2 - 7x        -2       3x          1
--------------- = ----- + --------- - -------
(x - 2)(x^2 +1)   x - 2    x^2 + 1    x^2 + 1

  x^2 - 7x        2       3x          1
--------------- = ----- + --------- - -------
(x - 2)(x^2 +1)   2 - x    x^2 + 1    x^2 + 1

-2 log (2 - x)



The integral comes out to

-ArcTan[x] - 2*Log[2 - x] + (3*Log[1 + x^2])/2

I cheated -- I used THE INTEGRATOR, at:

integrals.wolfram.com/index.jsp

But I had to 'fix it up' a bit, because this integral is undefined for  x <= 2.

Now if you look at the attached graph, you will see the problem -- is x = 1 the left boundary (I don't like the word 'limit' here.) or the right boundary?

If the problem says:

A curve has equation y = (7x-x^2) /((2-x) (x^2 +1)). Prove that the area of the region IN THE FIRST QUADRANT bounded by the curve, the x-axis, THE Y-AXIS, and the line x=1 is 7/2 ln2 - pi/4.

Then you use  x = 0, x = 1 as the boundaries.  [now you see why I had to fix it up.]

OK, let's do it:
................
-ArcTan[x] - 2*Log[2 - x] + (3*Log[1 + x^2])/2 (at x = 1)

= - arctan(1) - 2 ln(1) + 3 ln(2)/2

= - pi/4 + 3 ln(2)/2
..................
-ArcTan[x] - 2*Log[2 - x] + (3*Log[1 + x^2])/2 (at x = 0)

= - arctan(0) - 2 ln 2 + 3 ln(1)/2

=  0 - 2 ln(2) + 0

= - 2 ln 2

Final answer:

- pi/4 + 3 ln(2)/2 - (-2 ln 2)

- pi/4 + 3 ln(2)/2 + 2 ln 2

- pi/4 + 7 ln(2)/2

I guess that is what the book meant.  Time to throw that book away and get a good one.

AND, PLEASE, if you are going to write me again, remember -- you are not doing text-messaging, you are not doing IM-ing, you are writing me a letter.  Use proper grammar, spelling, etc. and proofread what you write.  You have the right to ask for an answer to your question.  You do not have the right to make me figure out what your question was.  If you expect me to take care in answering, you must take care in asking.