We have,
S=(x,y):3x2≤4y≤6x+24
Then,
y=43x2
4y=6x+24
Solving both equations, we get (−2,3) and (4,12).
Required area
A=∫−24(46x+24−43x2)dx
⇒A=41∫−24(6x+24−3x2)dx
⇒A=(43x2+6x−4x3)−24
⇒A=(12+24−16−3+12−2)
⇒A=27sq.units
JEE Main 2021 — Mathematics Calculus
The area of the region S=(x,y):3x2≤4y≤6x+24 is______.
Held on 26 Aug 2021 · Verified 6 Jul 2026.
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