# Area under a curve

• Jul 22nd 2010, 06:17 AM
Benji123
Area under a curve
Hello all!

Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

I would like to confirm whether the work I have done is correct or not
and if anyone could give me some guidance :)

∫3/x-2 dx - ∫2/x-1

3∫1/x-2 dx + 2∫1/x-1 dx

5
[3ln|x-2| + 2ln|x-1|]
3

(3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

= 4.682 units squared
• Jul 22nd 2010, 07:16 AM
Quote:

Originally Posted by Benji123
Hello all!

Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

I would like to confirm whether the work I have done is correct or not
and if anyone could give me some guidance :)

∫3/x-2 dx - ∫2/x-1

3∫1/x-2 dx + 2∫1/x-1 dx

5
[3ln|x-2| + 2ln|x-1|]
3

(3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

= 4.682 units squared

Looks good!

$\displaystyle \frac{3}{x-2}-\frac{2}{x-1}=\frac{3(x-1)-2(x-2)}{(x-2)(x-1)}=\frac{x+1}{(x-2)(x-1)}$

When applying the chain rule.....

$\displaystyle u=x-2\ \Rightarrow\ \frac{du}{dx}=1\ \Rightarrow\ du=dx$

$\displaystyle v=x-1\ \Rightarrow\ dv=dx$

hence all correct,

or is it?
• Jul 22nd 2010, 08:33 AM
Quote:

Originally Posted by Benji123
Hello all!

Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5

I would like to confirm whether the work I have done is correct or not
and if anyone could give me some guidance :)

∫3/x-2 dx - ∫2/x-1

3∫1/x-2 dx + 2∫1/x-1 dx

5
[3ln|x-2| + 2ln|x-1|]
3

(3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)

= 4.682 units squared

Unfortunately,

there is one error as you've calculated the integral of $\displaystyle \frac{3}{x-2}+\frac{2}{x-1}$

instead of the integral of $\displaystyle \frac{3}{x-2}-\frac{2}{x-1}$

so your calculated area should be less than 2.
• Jul 22nd 2010, 11:05 AM
Benji123
Therefore,

∫3/x-2 dx - ∫2/x-1

3∫1/x-2 dx - 2∫1/x-1 dx

5
[3ln|x-2| - 2ln|x-1|]
3

(3ln|5-2| - 2ln|5-1|) - (3ln|3-2| - 2ln|3-1|)

=0.523248144 - (-1.386294361)

= 1.910 units squared ?
• Jul 22nd 2010, 12:47 PM