Help Evaluating the Integral

[thatguy47]

Here's the problem:
\(\displaystyle \int_1^2 \frac{(4y^2 - 7y - 12)}{y(y +2)(y-3)} \, dy\)

I got \(\displaystyle \4y^2 - 7y - 12\) = A/(y+2) + B/(y-3) + C/y

After solving that I got (9/5)ln2 - (3/5)ln(4/3) which is incorrect. The answer is (27/5)ln2 - (9/5)ln3 or (9/5)ln(8/3) according to the book. How do you get to that?

 

[thatguy47]

Ya, sorry I can't make it look all pretty. Whats a good program I can buy/download to do that? Anyways....here's some of my work:

\(\displaystyle 4y^2 - 7y - 12\) = A(y-3)y + B(y+2)y + C(y+2)(y-3)
y=-2 >>> -6 = 10A >>> A= -.6
y=3 >>> 3 = 15B >>> B= .2
y=0 >>> -12= -6C >>> C= 2

From that I get -.6\(\displaystyle \int_1^2\)1/(y+2) + .2\(\displaystyle \int_1^2\)1/(y-3) + 2\(\displaystyle \int_1^2\)1/y

 

[Deleted member 4993]

Here's the problem:
\(\displaystyle \int_1^2 \frac{(4y^2 - 7y - 12)}{y(y +2)(y-3)} \, dy\)

I got \(\displaystyle \frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \, = \, \frac{A}{(y+2)}\, + \, \frac{B}{(y-3)}\, + \, \frac{C}{y}\)

After solving that I got (9/5)ln2 - (3/5)ln(4/3) which is incorrect. The answer is (27/5)ln2 - (9/5)ln3 or (9/5)ln(8/3) according to the book. How do you get to that?

You probably made mistake in calculating values of A, B & C.

Show your work - if you want us catch your mistake.

 

[Deleted member 4993]

A = (16+14-12)/(2*5) = 9/5

 

[galactus]

Here's a way to do partial fractions I prefer.

Multiply by the LCD....as you know.

\(\displaystyle Ay(y-3)+By(y+2)+C(y-3)(y+2)=4y^{2}-7y-12\)

Now, using each term, sub in what makes them 0. i.e. y-3=0, y=3

\(\displaystyle B(3)(3+2)=4(3)^{2}-7(3)-12\Rightarrow 15B=3\Rightarrow B=\frac{1}{5}\)

y+2=0, y=-2:

\(\displaystyle A(-2)(-2-3)=4(-2)^{2}-7(-2)-12\Rightarrow 10A=18\Rightarrow A=\frac{9}{5}\)

y=0:

\(\displaystyle C(0-3)(0+2)=-12\Rightarrow -6C=-12\Rightarrow C=2\)

This gives the PFD:

\(\displaystyle \frac{9}{5(y+2)}+\frac{1}{5(y-3)}+\frac{2}{y}\)

I thought I would show you this way of doing it in the event you had not seen it. It is easier than equating coefficients.

 

[thatguy47]

Thanks for the help guys. I got (9/5)ln(4/3) + (9/5)ln2 which is the same as the answers in the back, (27/5)ln2 - (9/5)ln3 or (9/5)ln(8/3) , when I input them in the calculator.

 

[thatguy47]

One last question for today:
\(\displaystyle \int \,\)(5x^2 + 3x - 2)/(x^3 +2x^2)

I got 5x^2 + 3x - 2 = A/x^2 + B/(x+2) + C/x

>>> 5x^2 + 3x - 2 = A(x + 2) + B(x^2) + C(x)(x+2)

>>>>x=-2 >>>> B=3
>>>>x=0 >>>> A= -1

How can I get C?
I did x = 1 >>> 6 = 3A + B + 4C
6= -3 + 3 +4C
C = 3/2

This isn't right because the answer says C=2 . How do you get to that?

 

[galactus]

I thought I just showed you in my post about PFD's

 

[thatguy47]

I thought I just showed you in my post about PFD's

I don't understand. If I do it that way do I get:
C(x)(x+2) = 5x^2 + 3x -2
and if I use x=0 then I would get C(0)(2) ....which wouldn't work.

 

[Deleted member 4993]

One last question for today:
\(\displaystyle \int \,\)(5x^2 + 3x - 2)/(x^3 +2x^2)

I got 5x^2 + 3x - 2 = A/x^2 + B/(x+2) + C/x

>>> 5x^2 + 3x - 2 = A(x + 2) + B(x^2) + C(x)(x+2)

>>>>x=-2 >>>> B=3
>>>>x=0 >>>> A= -1

How can I get C?
I did x = 1 >>> 6 = 3A + B + 4C
6= -3 + 3 +4C
C = 3/2

This isn't right because the answer says C=2 . How do you get to that?

\(\displaystyle 5x^2 + 3x + 2 = -1(x+2) + 3x^2 + Cx^2 +2C\)

Equating terms:

\(\displaystyle 5 = C + 3\)

 

[thatguy47]

One last question for today:
\(\displaystyle \int \,\)(5x^2 + 3x - 2)/(x^3 +2x^2)

I got 5x^2 + 3x - 2 = A/x^2 + B/(x+2) + C/x

>>> 5x^2 + 3x - 2 = A(x + 2) + B(x^2) + C(x)(x+2)

>>>>x=-2 >>>> B=3
>>>>x=0 >>>> A= -1

How can I get C?
I did x = 1 >>> 6 = 3A + B + 4C
6= -3 + 3 +4C
C = 3/2

This isn't right because the answer says C=2 . How do you get to that?

\(\displaystyle 5x^2 + 3x + 2 = -1(x+2) + 3x^2 + Cx^2 +2C\)

Equating terms:

\(\displaystyle 5 = C + 3\)

I see.... I added (x)(x+2) to equal 4 when x=1 instead of 3....stupid mistake.