# Solution to Problem 28

Here's a solution to
problem 28 for
Alvirne's AP Calculus Class, submitted by Sae-Hee Williams.
I've also written a solution to the pizza
problem, the other problem on Alvirne's AP Calculus page.

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## Problem

by Sae-Hee Williams
Find the area of the region below the curve and above the x-axis for:

f(x) = 'quot(x^2 + x + 1, x^4 + 16*x^3 + 94*x^2 + 240*x + 225)

where x .gteq 0 .

## Solution

by Ka-Ping Yee
The hard part of this problem is separating out the expression into
partial fractions that can be integrated. First, we have to factor
the denominator. The constant term is 225, which can be factorized
into primes as 225 = 3^2 * 5^2 . So, i tried dividing
x + 3 into the denominator by long division, and discovered that

x^4 + 16*x^3 + 94*x^2 + 240*x + 225 =
(x + 3) * (x^3 + 13*x^2 + 55*x + 75).

The next thing i tried was dividing by x + 5 , and that worked
too, leaving me with
(x + 3) * (x + 5) * (x^2 + 8*x + 15) .

Finally the quadratic is easy to factor, and we arrive at
x^4 + 16*x^3 + 94*x^2 + 240*x + 225 =
(x + 3)^2 * (x + 5)^2 .

So, the original rational function can be written as a sum of four
fractions like this:
f(x) = 'quot(A, x + 3) + 'quot(B, (x+3)^2)
+ 'quot(C, x + 5) + 'quot(D, (x+5)^2),

where the coefficients A, B, C, D are chosen to make the
numerator come out to x^2 + x + 1 when everything is
multiplied out again. The numerator of the big fraction will be
A*(x+3)*(x+5)^2 + B*(x+5)^2 + C*(x+3)^2*(x+5) + D*(x+3)^2

which expands to
x^3*(A+C) + x^2*(13*A + B + 11*C + D) + x*(55*A + 10*B + 39*C + 16*D)
+ 1*(75*A + 25*B + 45*C + 9*D) = x^2 + x + 1.

Matching coefficients gives us the four linear equations
A + C = 0

13*A + B + 11*C + D = 1

55*A + 10*B + 39*C + 10*D = 1

25*A + 25*B + 45*C + 25*D = 1

which we need to solve. I won't bore you with the dirty work,
but the result after substitution and cancellation is
A = _3,

B = 'frac(7,4),

C = 3,

D = 'frac(21,4).

This means that we can write
f(x) = _3/(x+3) + 'frac(7,4)/(x+3)^2 + 3/(x+5) + 'frac(21,4)/(x+5)^2

which is finally something we can integrate:
'integ(f(x), x) = _3*'ln(x+3) - 'frac(7,4)/(x+3)
+ 3*'ln(x+5) - 'frac(21,4)/(x+5)

Notice that
x^2 + x + 1 has no real zeroes; thus f(x)
never reaches the x-axis, and we need to integrate to infinity.
So the area we want is
'lim(3*('ln(x+5)-'ln(x+3)) - 'frac(7,4)/(x+3) - 'frac(21,4)/(x+5), x .approach 'inf) -
( _3*'ln(3) - 'frac(7,4)/3 + 3*'ln(5) - 'frac(21,4)/5)

But the difference 'ln(x+5) - 'ln(x+3) is equivalent to
'ln((x+5)/(x+3)), which becomes 'ln(1) = 0 as
x .approach 'inf. The fractions also head for zero,
so everything inside the limit ends up at zero, and the result is just
49/30 + 3*'ln(3) - 3*'ln(5)
.

This is the desired area under the given curve.

copyright © by **Ping** (e-mail) updated Sun 22 Sep 1996 at 22:04 JST

since Sun 22 Sep 1996