Definite integration

maggiehodgson

I have got as far as the .? But now I’m at a loss. The mark scheme ignores the ln of negative numbers. Why?

Reply 1

mqb2766

Original post

by maggiehodgson

I have got as far as the .? But now I’m at a loss. The mark scheme ignores the ln of negative numbers. Why?

?no pic again. Try posting it in a reply or on imgur and link.

Reply 2

maggiehodgson

Original post

by mqb2766

?no pic again. Try posting it in a reply or on imgur and link.

Reply 3

mqb2766

Original post

by maggiehodgson

Ok, Ive not worked it through properly, but when youre integrating (simplified numerator)
1/(x+4) + 1/(x-2)
from -3 to 1, you should note that the terms have vertical asymptotes at -4 and 2. The domain youre integrating over is between those values so you should be ok. The indefinite integral of the previous is
ln(|x+4|) + ln(|x-2|) + c
where the +c isnt that important (for your question) but the abs around the log args are important. For the domain were interested in so -3 to 1, then |x+4| = x+4, but |x-2|=-(x-2). So put the log args abs in the definite integral before evaluating the limits,
ln(|x+4|) + ln(|x-2|)
or just use the two previous relationships and evaulate the definite for
ln(x+4) + ln(-(x-2))
from -3 to 1.

(edited 8 months ago)

Reply 4

maggiehodgson

Thank you. I hadn’t thought of that.

Quick Reply

Related discussions

Latest

Trending

Articles for you

$What is LaTeX?$

What is LaTeX?

How The Student Room is moderated

To keep The Student Room safe for everyone, we moderate posts that are added to the site.

Try out the app

Continue on web