Given:
y2=−4x+4
y=2x+2

Required area
=∫−42(44−y2−2y−2)dy
=∫−42(2−4y2−2y)dy
=[2y−12y3−4y2]−42
=[(4−128−1)−(−8+1264−416)]
=15−6
=9 sq. units
The area enclosed between the curves y2+4x=4 and y−2x=2 is
Held on 24 Jan 2023 · Verified 6 Jul 2026.
325
322
9
323
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