Thread: 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

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!




When applying the chain rule.....





hence all correct,

or is it?

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

instead of the integral of

so your calculated area should be less than 2.

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 ?

Yes,
that's it!

Thanks I really appreciate it and pointing out my error. I don't get why alot of revision notes don't give you answers to the questions so you can check.