integration area help
Maths and statistics discussion, revision, exam and homework help.
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integration area help
find the area of the shades region; y=x^2-4
the first area is where x is 0 and x is 2
second area where x is 2 and x is 3
first area is underneath x axis, second area above x axis
so i did limits and got -16/3 for first area ( i know you just make this positive) but for the second area which is above the x axis i got -41/3? how can this be? someone help -
Re: integration area helpwell first area i got -16/3 but as its underneath x-axis you just count it as positive so 16/3. second area i got -41/3 but it is above x axis. I should add both numbers together but surely im wrong if the area above the x axis is a negative? so either 19 or -25/3 lol. what did you get can you help(Original post by Sketch)
Can i ask what the answer to the question is? -
Re: integration area help-3(Original post by sputum)
If you run x=3 through your integral what do you get? -
Re: integration area helpKeerect!(Original post by dongonaeatu)
-3
Now take what you got for x=2 away from this. -
Re: integration area helpsecond area i did(Original post by Sketch)
hmmm, for the second area i got 7/3.. are you sure your method is correct? talk me through your working out for the second area..
[(3)^3/3-4(3)] - [(2)^3/3-4(2)]= -41/3
wait i just did it again... i got 7/3 this time
wtf it will still end up being negative -3+7/3= -2/3
oh wait.. the whole second area is 7/3 so its 16/3+7/3= 23/3 for the whole area?Last edited by dongonaeatu; 13-04-2012 at 17:12. -
Re: integration area helpyes!(Original post by dongonaeatu)
the answer for the whole area is 23/3 yes or no -
Re: integration area helpputting x=3 in y=x3/3 -4x(Original post by Sketch)
where did the -3 come from may i ask?