Quote:
Originally Posted by hawk97
The first step for evaluating the integral is to break up the fraction so that the denominator is simpler.
The denominator can be factored into (2x-5)(x-2) so …
5
⌠ ( 4x - 9 ) * dx
⌡ (2x-5)(x-2)
3
This fraction can now be broken in half…
A/(2x-5) + B/(x-2)
A and B have taken the place of the numerator, because we do not know what their values are. To find their values we set the original fraction equal to A/(2x-5) + B/(x-2).
( 4x - 9 )/((2x-5)(x-2)) = A/(2x-5) + B/(x-2)
4x-9 = A(x-2) + B(2x-5)
Then you substitute in numbers for x, because this equation should be true for all numbers. So…
If x =2
-1 = -1*B
B=1
If x = 5/2
1 = ˝ *A
A = 2
Now we return to our split-up fraction and substitute them in
5
∫ (A/(2x-5) + B/(x-2))*dx =
3
5
∫ (2/(2x-5) + 3/(x-2))*dx
3
Now we can proceed with taking the integral…
5
∫ (2/(2x-5) + 3/(x-2))*dx =
3
We will solve this in two parts
5
∫ (2/(2x-5))*dx
3
u = 2x-5
du = 2dx
dx = du/2
so we substitute in u for 2x-5 and du/2 for dx …
∫ (2/u)*du/2 =
∫ (1/u)*du = ln(u)
We then substitute the 2x-5 back in…
ln (2x-5) from 3 to 5 =
ln (10-5) – ln (6-5) = ln(5) – ln(1)
now we solve the other half…
∫dx/(x-2)
u = x-2
du = dx
∫du/u = ln(u)
ln(u) = ln(x-2)
ln(x-2) from 3 to 5 = ln(5-2) – ln(3-2) = ln(3) – ln(1)
now we add the two halves together
ln(5) – ln(1) + ln(3) – ln(1)
at some point you need to remember that ln(1) = 0 
so you get ………. Answer choice A) ln(5)+ln(3)  |
The solution you suggest is correct but there is a quicker way to solve this:
If you notice the (4x-9) is the derivative (I don;t know if this is the right word) of the 2(x^2)-9x+10.
that is to say (2(x^2)-9x+10)' =4x-9
So,the only thing you have to solve is: (ln2(x^2)-9x+10)^5 - (ln2(x^2)-9x+10)^3 and if you solve it you will find ln(15)=ln(3*5)=ln(3)+ln(5)
04-06-2008
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