Sorry, another integration question. (1 Viewer)

[RHINO7]

Find the area enclosed between the curve y= 2/(x-3)^2, the x-axis and the lines x=0 and x=1

 

[Sloppy Sailor]

I don't know how to use the integrate sign here!

bounds: a = 0, b = 1

= 2 (x-3) ^-2
= [2 (x-3) ^ -1 / -1]
= [-2 / (x-3)] --> sub in the bounds the normal way
= 1/2 - 2/3
= -1/6

A = | -1/6 |
= 1/6 u^2

Maybe.. I don't like the negative area :| I dont have a calculator on me so I probably made a mistake somewhere.. anyway, hope that kinda helped.

wait, wtf. Shouldn't it be [(2 x (1-3)^-1/ -1) - (2 x (0-3)^-1/ -1)] = [(1) - (2/3)] = 1/3

 

[ianc]

wait, wtf. Shouldn't it be [(2 x (1-3)^-1/ -1) - (2 x (0-3)^-1/ -1)] = [(1) - (2/3)] = 1/3

Yes thats what it should be. theres a lot of minuses floating round there so you've got to be really careful when substituting in the limits.

also with this question, because it was asking for area, its a good idea to sketch it in case some of it goes below the x axis (but in this case you didnt need to worry about splitting up the area into 2 different integrals)