ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: abd123 on December 09, 2011, 09:27:57 am

Title: A123 Maths Bonanza.
Post by: abd123 on December 09, 2011, 09:27:57 am: This question is a bit strange once, i got stuck on it yesterday night and tried to have a go again in the morning.

$\int_{-1}^{1}\frac{-1}{4-(x-1)^2}dx$
$I=\int_{-1}^{1}\frac{1}{(x-1)^2-4}dx$
$=\int_{-1}^{1}\frac{1}{(x-3)(x+1)}dx$

$=\frac{1}{4}\int_{-1}^{1}\frac{1}{x-3}-\frac{1}}{x+1}dx$
$=\frac{1}{4} [log_e\vert\frac{x-3}{x+1}\vert]_{-1}^{1}dx$
$=\frac{1}{4}(log_e(1)-log_e(\frac{4}{0})$

Helps Appreciated :)
Title: Re: A123 Maths Bonanza.
Post by: dc302 on December 09, 2011, 10:17:06 am: Question must be wrong. What does the answer say?
Title: Re: A123 Maths Bonanza.
Post by: ArchieMae on December 09, 2011, 10:41:20 am: Try doing the partial fractions without first multiplying by -1 as you did in step 1.
Title: Re: A123 Maths Bonanza.
Post by: xZero on December 09, 2011, 12:19:37 pm: At x=-1, the graph has an asymptote so you can't integrate it at that point.