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JEE Main 2024 · Mathematics · Area Under Curves · hard

Let A be the area, in square units, of the region lying above the curve y=|x^2-4| and below the line y=5. The value of 3A is

Numerical answer

Let AA be the area, in square units, of the region lying above the curve y=∣x2−4∣y=|x^2-4| and below the line y=5y=5. The value of 3A3A is
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Answer: 44

The line meets y=∣x2−4∣y=|x^2-4| where ∣x2−4∣=5|x^2-4|=5, i.e. x2=9x^2=9 (the case x2−4=−5x^2-4=-5 has no real root), so x=±3x=\pm3. The region is symmetric about the yy-axis. For 0≤x≤20\le x\le2 the curve is y=4−x2y=4-x^2, so the vertical gap is 1+x21+x^2; for 2≤x≤32\le x\le3 it is y=x2−4y=x^2-4, so the gap is 9−x29-x^2. Then A2=∫02(1+x2)dx+∫23(9−x2)dx=(2+83)+(9−193)=223\dfrac A2=\int_0^2(1+x^2)dx+\int_2^3(9-x^2)dx=\left(2+\dfrac83\right)+\left(9-\dfrac{19}3\right)=\dfrac{22}3. Hence A=443A=\dfrac{44}3 and 3A=443A=44. The line y=5y=5 lies above the curve for ∣x∣<3|x|<3 only, so the integration limits are −3-3 to 33.
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