integration (1 Viewer)

[ashokkumar]

find the definite integral of (7+x-2x^2)/(2-x) from 1 to 0.

 

[kcqn93]

use long division
to get:
(7+x-2x^2)/(2-x) = 2x+3 + 1/(2-x)

then intergrate

i got 4 + ln2

not too sure... am i right?

 

[ashokkumar]

yea that's right. but when i integrated and substituted 1 and 0, i ended up with ln(-1) and ln(-2) as part of my answer. I took the absolute values inside the bracket and simplified, ending up with the correct answer.


Can i do this?

 

[kcqn93]

you shouldn't get any negative values inside the logs

i got



from 0 to 1 i.e sub in 1 first

 

[ashokkumar]

oooh. at the start of the question, i multiplied the numerator and denominator by -1 because i wanted the leading terms of each to be positive :S

As a result, i got integral from 1 to 0 of (2x + 3) -1/(x+2)

 

[cutemouse]

Yes, you're allowed to do that.

Consider the graph of y=1/x. Say you want to find the area between x= - 2 and x = -1 then by symmetry this is the same as the area between x=1 and x=2.

Or just use absolute values (easier )

 

[Trebla]

yea that's right. but when i integrated and substituted 1 and 0, i ended up with ln(-1) and ln(-2) as part of my answer. I took the absolute values inside the bracket and simplified, ending up with the correct answer.


Can i do this?

The way your answer is written something of say the form:
ln(-1) - ln(-2) should really be "ln(-2 / -1) = ln 2"
because the final answer is the combination of the logs which is well defined.

Basically you should always try to combine all your logarithmic functions in the final answer as soon as you integrated rather than keep them separate. In this course, you should never end up with a stand alone log of a negative number. The logs of negative numbers often come in addition/subtraction pairs which can be combined to give a valid answer. By combining them as soon as possible, should avoid the confusion of invalid results.

 

[cutemouse]

Or just use absolute values... lol

 

[Affinity]

One reminder .. the integral of 1/x is (ln |x|) not (ln(x))

 

[kcqn93]

One reminder .. the integral of 1/x is (ln |x|) not (ln(x))

oooohh i didnt know that! :S haha