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1. 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
2. Re: integration area help
3. Re: integration area help
(Original post by Sketch)
well 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
4. Re: integration area help
If you run x=3 through your integral what do you get?
5. Re: integration area help
(Original post by sputum)
If you run x=3 through your integral what do you get?
-3
6. Re: integration area help
(Original post by dongonaeatu)
-3
Keerect!
Now take what you got for x=2 away from this.
7. Re: integration area help
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..
8. Re: integration area help
oh.. so is it meant to be -3--41/3? with 2 minus's making a plus
9. Re: integration area help
(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..
second area i did

[(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.
10. Re: integration area help
7/3 is good
11. Re: integration area help
The area in this case would be the integral of the "above x axis" area, minus MINUS the integral of the "below x axis" area.. i.e. just take the sum of their absolute values..
12. Re: integration area help
the answer for the whole area is 23/3 yes or no
13. Re: integration area help
where did the -3 come from may i ask?
14. Re: integration area help
(Original post by dongonaeatu)
the answer for the whole area is 23/3 yes or no
yes!
15. Re: integration area help
(Original post by Sketch)
where did the -3 come from may i ask?
putting x=3 in y=x3/3 -4x
16. Re: integration area help
17. Re: integration area help
uh ran out of rep..sputnam i'll rep you later bro thanks