Area Bounded questions

[anraymer]

Okay, I have this problem:
Find the area bounded by y= (x-3)/(x^3+x^2), y=0, x=1, x=2

:?:
I know you have to simplify then use integration to get the answer but I am so confused and can't seem to solve this one.


Any ideas?

 

[galactus]

You could expand:

\(\displaystyle \L\\\frac{x-3}{x^{3}+x^{2}}=\frac{4}{x}-\frac{3}{x^{2}}-\frac{4}{x+1}\)

Looks like you have some "ln's" when you integrate.

Or, just use the quotient rule. There are variuos ways to tackle it.

 

[soroban]

Hello, anraymer!

Find the area bounded by: \(\displaystyle \:y\:=\:\frac{x\,-\,3}{x^3\,+\,x^2}\;\; y\,=\,0,\;x\,=\,1,\; x\,=\,2\)

Exactly where is your difficulty?
. . You don't know that integration gives us area?
. . You can't integrate that function?
. . You don't know how to evaluate definite integrals?

We have: \(\displaystyle \L\:\int^{\;\;\;2}_1\frac{x\,-\,3}{x^2(x\,+\,1)}\,dx\)

Partial fractions: \(\displaystyle \L\:\frac{x\,-\,3}{x^2(x\,+\,1)} \;=\;\frac{4}{x}\,-\,\frac{3}{x^2}\,-\,\frac{4}{x\,+\,1}\)

Can you finish it now?

The area is entirely below the x-axis,
. . so the answer comes out negative.