Integration question (1 Viewer)

qUiCkMaThS

New Member
Joined
Aug 6, 2020
Messages
6
Gender
Undisclosed
HSC
2019
well, idk what I've done wrong, but if you find it, just point it out:

A1 + A2 + A3 = ATotal
A2 + A3 = 5A1/ 2

Since A1 = 3 x 2 = 6

ATotal = 15+6 = 21

i got 21 aswell, but I thought we subtract the bottom two from the area of the top? not sure tho
I've forgotten whether 'evaluate' changes what the question is asking....sorry about that!
 
Joined
Jan 5, 2021
Messages
106
Gender
Undisclosed
HSC
2021
well, idk what I've done wrong, but if you find it, just point it out:

A1 + A2 + A3 = ATotal
A2 + A3 = 5A1/ 2

Since A1 = 3 x 2 = 6

ATotal = 15+6 = 21



I've forgotten whether 'evaluate' changes what the question is asking....sorry about that!
no problem!! that's exactly what I got, it may be an error in the question, thank you!!
 

B1andB2

oui oui baguette
Joined
Nov 6, 2019
Messages
576
Location
cuddles lane
Gender
Undisclosed
HSC
N/A
A1=6
A2= 3
A3 = 5(3x2)-2(3)/2 = 12

when you evaluate the integral (not the same as finding area) you consider the fact that below= negative

so solving the integral = -6+3-12 = -15
 
Joined
Jan 5, 2021
Messages
106
Gender
Undisclosed
HSC
2021
A1=6
A2= 3
A3 = 5(3x2)-2(3)/2 = 12

when you evaluate the integral (not the same as finding area) you consider the fact that below= negative

so solving the integral = -6+3-12 =15
yesss that makes sense, thank you tooo!!
 

YonOra

Well-Known Member
Joined
Apr 21, 2020
Messages
375
Gender
Undisclosed
HSC
2021
A1=6
A2= 3
A3 = 5(3x2)-2(3)/2 = 12

when you evaluate the integral (not the same as finding area) you consider the fact that below= negative

so solving the integral = -6+3-12 = -15
How'd u get A2 to be 3. Having an elderly moment
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top