Area between curves (1 Viewer)

Drewk · Feb 28, 2013

http://snag.gy/9ooiP.jpg
With the question above i tried to use integration from 0 to 1 then multplied by 2
But i dont get 2/3 units ??
Please help

Nws m8 · Feb 28, 2013

Edit: it is 0 to 1 for (x-1)^2

It could also be -1 to 0 for (x+1)^2

Drongoski · Feb 28, 2013

Required area is:

$2 \times \int _0 ^1 (x-1)^2 dx = 2 \left [ \frac {(x-1)^3}{3} \right ]^1 _0 = \frac {2}{3}$

Drewk · Feb 28, 2013

Nws m8 said:

Why have u only integrated one of the functions i thought we have to take both functions and go TOP-BOTTOM etc.
Thanks

Nws m8 · Feb 28, 2013

Drewk said:
Why have u only integrated one of the functions i thought we have to take both functions and go TOP-BOTTOM etc.
Thanks

By symmetry we know the two regions have the same area, so we can just find one area using any equation with CORRECT limits and times it by 2

iSplicer · Feb 28, 2013

Nws m8 said:
By symmetry we know the two regions have the same area, so we can just find one area using any equation with CORRECT limits and times it by 2

Make sure you actaully use the symmetry argument in the exam to ensure full marks!!

Area between curves (1 Viewer)

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