ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lacoste on November 02, 2009, 03:52:55 pm

Title: why?
Post by: lacoste on November 02, 2009, 03:52:55 pm: (http://i34.tinypic.com/2l9jz37.jpg)

Thanks~! :)
Title: Re: why?
Post by: naved_s9994 on November 02, 2009, 04:00:28 pm: h(x) seems to be 1/3x
you can work it out from there?
Title: Re: why?
Post by: m@tty on November 02, 2009, 04:37:36 pm: You need to split the second integral up.
$= 2 \int_{5}^{1} h(x)\ dx - \int_{5}^{1} 2\ dx$
$= -2 \int_{1}^{5} h(x)\ dx + \int_{1}^{5} 2\ dx$

From there you should be able to work it out, with the given definite integral.

$= -2(4)\ + [2x]_{1}^{5}$
$= -2(4) + 2(5)-2(1) = 0$