A small MX2 question (1 Viewer)

[Mrmellow]

Hello guys and gals. Okay, so I was going through some MX2 work and got thumped by a question. This is probably a stupid question but anyway. So the question asked to find the area between the curve y= (3x^2 -7)/(x-2)(x+3) and the x-axis. Now, the question earlier had asked to show that (1/x-2)- (4/x+3)+3 was the same as the above expression, which I showed and later used in the question. To find the area I tried the integral of the substitute expression which came out as [ln(x-2)-4ln(x+3)+3]from sqrt(7/3) to -sqrt(7/3). Plugging in the numbers I found it to be undefined, but the answer at the back said 2.66 units^2 and so did wolfram alpha. Any help in this issue would be appreciated.

 

[dunjaaa]

Try ln|x-2|-4ln|x+3| + 3 (Absolute value brackets). We use them in integration because the log function cannot be less than or equal to 0.

 

[Luckytree]

uhm to avoid the undefined answer use absolute value sign for ln(x-2). i.e. ln(|x-2|)

 

[Luckytree]

beat me to it HAHA

 

[Mrmellow]

Makes sense now, thanks everyone

 

[Mrmellow]

So, um, is this just a nifty trick that you ought to know when doing these problems?

 

[Luckytree]

i forgot the reason my teacher told me, but esp in mechanics (resisted motion) we take into account direction and hence we do the absolute value thingo
this reason could be wrong, meh.

 

[faisalabdul16]

So, um, is this just a nifty trick that you ought to know when doing these problems?

When you integrate something and get ln something, you always put absolute value signs if there is a bound.

Forgot the reasoning, however i don't think you don't need to know, just remember to do it.

 

[Mrmellow]

I thought you only put the abs value sign out side the [] box thing and not with in the function, i.e. ln|x-2|. If you know what I mean...

 

[dunjaaa]

Functions like the log function have conditions on the domain (i.e. x>0). That's why we absolute the inside, instead of the outside.