Question

Draw a rough sketch of the curve, $y=\frac{x^2+3x+2}{x^2-3x+2}$ and find the area of the bounded region between the curve and x-axis.

Answer 1

$f\left(x\right)=\frac{(x+1)(x+2)}{(x-1)(x-2)}$

Graph will cut x-axis at $x=-1$ and $x=-2$

It is discontinuous at $x=1$ and $x=2$.
$\underset{x\to \infty }{lim}f\left(x\right)\to 1,\underset{x\to 1^{-}}{lim}f\left(x\right)\to +\infty$

$\underset{x\to 1^{+}}{lim}f\left(x\right)\to -\infty$

$\underset{x\to 2^{-}}{lim}f\left(x\right)\to -\infty$

$\underset{x\to 2^{+}}{lim}f\left(x\right)\to +\infty ,f\left(0\right)=1$.

Now we have to find the area of the shaded region. The required area

$=\left|{\int }_{-2}^{-1}f\right(x\left)dx\right|=\left|{\int }_{-2}^{-1}\left(\frac{x^2+3x+2}{x^2-3x+2}\right)dx\right|=|{\int }_{-2}^{-1}1+\frac{6x}{(x-1)(x-2)}dx|$

$=\left|[x{|}_{-2}^{-1}+6{\int }_{-2}^{-1}(\frac{2}{x-2}-\frac{1}{x-1})dx]\right|$

$=|[1+6[2ln|x-2|-ln|x-1|{]}_{-2}^{-1}|$

$=|1+6\left[2\right(ln3-ln4)-(ln2-ln3\left)\right]|$

$=|1+6[3ln3-5ln2]|$

$=6ln\left(\frac{32}{27}\right)-1$ sq. units.