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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Find the area of the region below the curve and above the x-axis for:
wheref(x) = 'quot(x^2 + x + 1, x^4 + 16*x^3 + 94*x^2 + 240*x + 225)
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
The next thing i tried was dividing byx^4 + 16*x^3 + 94*x^2 + 240*x + 225 = (x + 3) * (x^3 + 13*x^2 + 55*x + 75) .
Finally the quadratic is easy to factor, and we arrive at(x + 3) * (x + 5) * (x^2 + 8*x + 15) .
So, the original rational function can be written as a sum of four fractions like this:x^4 + 16*x^3 + 94*x^2 + 240*x + 225 = (x + 3)^2 * (x + 5)^2 .
where the coefficientsf(x) = 'quot(A, x + 3) + 'quot(B, (x+3)^2) + 'quot(C, x + 5) + 'quot(D, (x+5)^2) ,
which expands toA*(x+3)*(x+5)^2 + B*(x+5)^2 + C*(x+3)^2*(x+5) + D*(x+3)^2
Matching coefficients gives us the four linear equationsx^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 .
which we need to solve. I won't bore you with the dirty work, but the result after substitution and cancellation isA + 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
This means that we can writeA = _3 ,
B = 'frac(7,4) ,
C = 3 ,
D = 'frac(21,4) .
which is finally something we can integrate:f(x) = _3/(x+3) + 'frac(7,4)/(x+3)^2 + 3/(x+5) + 'frac(21,4)/(x+5)^2
Notice that'integ(f(x), x) = _3*'ln(x+3) - 'frac(7,4)/(x+3) + 3*'ln(x+5) - 'frac(21,4)/(x+5)
But the difference'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)
49/30 + 3*'ln(3) - 3*'ln(5) .
This is the desired area under the given curve.