Application of IntegralsmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Region Bounded by Line and Parabola: Area = 9 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The area of the region {(x,y):0≤y≤6−x, y2≥4x−3, x≥0}\{(x,y):0\le y\le6-x,\ y^2\ge4x-3,\ x\ge0\} is
A88
B99correct
C1212
D1515
Solution
Step 1: y2≥4x−3⇒x≤y2+34y^2\ge4x-3\Rightarrow x\le\dfrac{y^2+3}{4}; 0≤y≤6−x⇒x≤6−y, y≥00\le y\le6-x\Rightarrow x\le6-y,\ y\ge0; with x≥0x\ge0, each strip: x=0x=0 to x=min⁡(y2+34, 6−y)x=\min\left(\dfrac{y^2+3}{4},\ 6-y\right). Step 2: y2+34=6−y⇒y2+3=24−4y⇒y2+4y−21=0⇒(y+7)(y−3)=0⇒y=3\dfrac{y^2+3}{4}=6-y\Rightarrow y^2+3=24-4y\Rightarrow y^2+4y-21=0\Rightarrow(y+7)(y-3)=0\Rightarrow y=3. For 0≤y≤30\le y\le3: parabola nearer; for 3≤y≤63\le y\le6: line nearer. Step 3:
A1=∫03y2+34 dy=14[y33+3y]03=14(9+9)=92.A_1=\int_0^3\dfrac{y^2+3}{4}\,dy=\dfrac14\left[\dfrac{y^3}{3}+3y\right]_0^3=\dfrac14\left(9+9\right)=\dfrac92.
 Step 4:
A2=∫36(6−y) dy=[6y−y22]36=(36−18)−(18−92)=18−272=92.A_2=\int_3^6(6-y)\,dy=\left[6y-\dfrac{y^2}{2}\right]_3^6=(36-18)-\left(18-\dfrac92\right)=18-\dfrac{27}{2}=\dfrac{9}{2}.
 Step 5:
∴A=A1+A2=92+92=9.\therefore A=A_1+A_2=\dfrac92+\dfrac92=9.
 Correct answer: (2)
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