You have to split the integral up as .
I'm not sure how the (x+1) part would cause trouble. You should get .
Hi, I would be very grateful if anyone could help me with this problem. I can pretty much see how you're meant to do this, but one bit is causing a large problem, the result when you integrate the (x+1) function, in the answer given was rather vague as to why it behaved strangely at this point. Cheers!
Hello LHS
So the lower limit of the integral lies between and inclusive, and the upper limit between and inclusive. So we shall need to split the integral into two ranges: to and to , to take into account the change in the function where . This gives:
All you need to do now is to integrate each of these functions separately and evaluate them between their respective limits in the usual way, before simplifying the resulting polynomial in .
You asked specifically about the function. This integrates to , of course, and when you put in the limits you'll get:
When you combine this with the result of the second integral, I think the final answer is:
Can you fill in the gaps?
Grandad
Thank you both of you, that confirms the method I was using. I asked about that specific integral because I was confused why one ended up with an extra 0.5, but now I see that this cancels when you use put it between the limits... there must be a slip in my workings somewhere..
Thank you very much for your in-depth responses