Given curves are y=2{x}^{2}&y=4-2x
So, 2x2=4−2x
⇒2x2+2x−4=0⇒(x−1)(x+2)=0
So, the x-coordinate of the intersection point is 1 as x=−2<0
Required area ∫01(4−2x−2x2)dx=[4x−x2−32x3]01
=4−1−32=37
The area (in sq. units) of the region, given by the set (x,y)∈R×R∣x≥0,2x2≤y≤4−2x is :
Held on 25 Jul 2021 · Verified 6 Jul 2026.
38
317
313
37
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