# I need help with Integration question

#1
0
3 months ago
#2
(Original post by Matheen1)
As usual, what have you done?
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#3
(Original post by mqb2766)
As usual, what have you done?
0
3 months ago
#4
(Original post by Matheen1)
Thats not quite what the question is asking.
Sketch the curve and the boundary of the integration problem and upload?
0
#5
(Original post by mqb2766)
Thats not quite what the question is asking.
Sketch the curve and the boundary of the integration problem and upload?
0
3 months ago
#6
Which part are you working out?
Also note your previous working misses out the lower limit part at the end.
Last edited by mqb2766; 3 months ago
0
#7
I’m trying to find the enclosed area in the middle.
0
3 months ago
#8
(Original post by Matheen1)
I’m trying to find the enclosed area in the middle.
Agreed, you've only done from x = 1/2 .. 2 though, you're missing a bit.
Also note that your working doesn't include the lower or upper(?) limit in the previous post. The area should not be negative.
Last edited by mqb2766; 3 months ago
0
#9
(Original post by mqb2766)
Agreed, you've only done from x = 1/2 .. 2 though, you're missing a bit.
Also note that your working doesn't include the lower or upper(?) limit in the previous post. The area should not be negative.
0
#10
(Original post by Matheen1)
I really don’t know what to do.
0
3 months ago
#11
(Original post by Matheen1)
I really don’t know what to do.
3/2 looks about right for 1/2..2. So what about the bit from 0..1/2.
Id bet you do know how to get that area.
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#12
(Original post by mqb2766)
3/2 looks about right for 1/2..2. So what about the bit from 0..1/2.
Id bet you do know how to get that area.
Wait how did u get 0…1/2?
0
3 months ago
#13
(Original post by Matheen1)
Wait how did u get 0…1/2?
You want to find the area enclosed by the axes, the curve and x=2 and y=4. Do a sketch, rather than desmos, and shade in the area youve already calculated. Which bit is missing?
0
#14
(Original post by mqb2766)
You want to find the area enclosed by the axes, the curve and x=2 and y=4. Do a sketch, rather than desmos, and shade in the area youve already calculated. Which bit is missing?
Isn’t the part that is left meant to be 0…2 I just can’t get my head around why it is 0…1/2
0
3 months ago
#15
(Original post by Matheen1)
Isn’t the part that is left meant to be 0…2 I just can’t get my head around why it is 0…1/2
Tbh, Id forget about desmos (its good, but just sketch it on your paper as you would do in an exam).
Shade the area youve calculated, noting the limits youve used.
Last edited by mqb2766; 3 months ago
0
3 months ago
#16
Wouldn't this be the correct way to draw the graph (R = area)?

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