Integration question help (1 Viewer)

[Marcus Aurelius]

Hey guys This was a question in my exam and while the working was good I got it wrong because one of the signs is meant to be positive and i had it as a negative but i don't know why that is. I'm not sure what the answer is but its not 3.5 cause thats what i got. please help its getin annoying because there is probably a very simple explanation. Here it is I hope it makes sense :
btw x is not times

Evaluate
(Integration sign)1-2 2x - 1/xsquared dx

Sorry i don't know how to make the correct maths symbols on my keyboard
Thanx

 

[Pace_T]

(Integration sign)1-2 2x - 1/xsquared dx

= S (2x/x^2 - 1/x^2) 1 - 2
= S ( 2/x - 1/x^2) 1 - 2
= (2ln(x) + 1/x) 1 - 2
= 2ln2 + 1/2 - (2ln1 + 1)
= 0.8862

 

[Slidey]

For future reference:
^ means "to the power of", so x square is: x^2

Two x minus 1 all on x squared is (2x-1)/x^2 otherwise it is ambiguous as 2x-1/x^2 and could be something else.

 

[Marcus Aurelius]

Ah my question could be misinterpreted. Here it is again with added brackets so its not read as 2x-1 above x squared. My mistake

Evaluate
(Integration sign)1-2 (2x) - (1/x^2)

 

[Slidey]

Ahhh. No, sorry. You didn't need extra brackets. Sometimes people do and it's confusing. Pace's solution is for another situation.

Integral (2x - 1/x^2) from 1 to 2
Break up:
Integral of 2x - Integral of x^-2 both from 1 to 2:
x^2+1/x from 1 to 2
(4+1/2)-(1+1)
2+1/2=2.5 #

 

[Pace_T]

sorry, my bad

 

[Slidey]

If it's any consolation, I made the same mistake.

 

[Marcus Aurelius]

Yes thats the correct answer cause i remember i was one off with my 3.5. i made the mistake with the 1/x^2 part and had the wrong signs, neg where is was supposed to be pos. Alright cool, thanks for the help guys.