Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Area Under the Curves question

2025 · 28 Jan · Shift 1 · Q45
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Area Under the Curves
  5. /2025 · 28 Jan · Shift 1 · Q45

Area Under the Curves question

2025 · 28 Jan · Shift 1 · Q45

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region {(x,y):0≤y≤2∣x∣+1,0≤y≤x2+1,∣x∣≤3}\left\{(x, \mathrm{y}): 0 \leq \mathrm{y} \leq 2|x|+1,0 \leq \mathrm{y} \leq x^2+1,|x| \leq 3\right\}{(x,y):0≤y≤2∣x∣+1,0≤y≤x2+1,∣x∣≤3} is
  1. A
    323\frac{32}{3}332​
  2. B
    643\frac{64}{3}364​
  3. C
    173\frac{17}{3}317​
  4. D
    803\frac{80}{3}380​
View written solutionFree

Correct answer: B

  1. Interpret the region

The region is

{(x,y):0≤y≤2∣x∣+1,  0≤y≤x2+1,  ∣x∣≤3}.\{(x,y): 0\le y\le 2|x|+1,\; 0\le y\le x^2+1,\; |x|\le 3\}.{(x,y):0≤y≤2∣x∣+1,0≤y≤x2+1,∣x∣≤3}.

So for each fixed xxx with −3≤x≤3-3\le x\le 3−3≤x≤3, yyy must satisfy both y≤2∣x∣+1andy≤x2+1.y\le 2|x|+1 \quad \text{and} \quad y\le x^2+1.y≤2∣x∣+1andy≤x2+1. Hence the upper boundary is y=min⁡(2∣x∣+1,  x2+1).y=\min(2|x|+1,\; x^2+1).y=min(2∣x∣+1,x2+1).

Therefore the required area is

∫−33min⁡(2∣x∣+1,  x2+1) dx.\int_{-3}^{3} \min(2|x|+1,\; x^2+1)\,dx.∫−33​min(2∣x∣+1,x2+1)dx.
  1. Find where the two curves intersect

We solve x2+1=2∣x∣+1.x^2+1=2|x|+1.x2+1=2∣x∣+1. This gives x2=2∣x∣.x^2=2|x|.x2=2∣x∣. Let t=∣x∣≥0t=|x|\ge 0t=∣x∣≥0. Then t2=2t  ⟹  t(t−2)=0,t^2=2t \implies t(t-2)=0,t2=2t⟹t(t−2)=0, so

Hence intersection points occur at x=0,  x=±2.x=0,\; x=\pm 2.x=0,x=±2.

  1. Decide which curve is lower

We compare x2+1and2∣x∣+1.x^2+1 \quad \text{and} \quad 2|x|+1.x2+1and2∣x∣+1. Subtracting 111 from both, compare x2x^2x2 and 2∣x∣2|x|2∣x∣.

  • If ∣x∣<2|x|<2∣x∣<2, then x2<2∣x∣x^2<2|x|x2<2∣x∣, so x2+1<2∣x∣+1.x^2+1<2|x|+1.x2+1<2∣x∣+1. Thus the lower curve is y=x2+1y=x^2+1y=x2+1.

  • If 2<∣x∣≤32<|x|\le 32<∣x∣≤3, then x2>2∣x∣x^2>2|x|x2>2∣x∣, so x2+1>2∣x∣+1.x^2+1>2|x|+1.x2+1>2∣x∣+1. Thus the lower curve is y=2∣x∣+1y=2|x|+1y=2∣x∣+1.

So,

Area=∫−22(x2+1) dx+∫[−3,−2]∪[2,3](2∣x∣+1) dx.\text{Area} = \int_{-2}^{2} (x^2+1)\,dx + \int_{[-3,-2]\cup[2,3]} (2|x|+1)\,dx.Area=∫−22​(x2+1)dx+∫[−3,−2]∪[2,3]​(2∣x∣+1)dx.

Using symmetry,

Area=2[∫02(x2+1) dx+∫23(2x+1) dx].\text{Area} = 2\left[\int_0^2 (x^2+1)\,dx + \int_2^3 (2x+1)\,dx\right].Area=2[∫02​(x2+1)dx+∫23​(2x+1)dx].
  1. Evaluate the integrals

First,

∫02(x2+1) dx=[x33+x]02=83+2=143.\int_0^2 (x^2+1)\,dx = \left[\frac{x^3}{3}+x\right]_0^2 = \frac{8}{3}+2 = \frac{14}{3}.∫02​(x2+1)dx=[3x3​+x]02​=38​+2=314​.

Next,

∫23(2x+1) dx=[x2+x]23=(9+3)−(4+2)=6.\int_2^3 (2x+1)\,dx = \left[x^2+x\right]_2^3 = (9+3)-(4+2)=6.∫23​(2x+1)dx=[x2+x]23​=(9+3)−(4+2)=6.

Therefore,

Area=2(143+6)=2(143+183)=2⋅323=643.\text{Area} = 2\left(\frac{14}{3}+6\right) = 2\left(\frac{14}{3}+\frac{18}{3}\right) = 2\cdot \frac{32}{3} = \frac{64}{3}.Area=2(314​+6)=2(314​+318​)=2⋅332​=364​.
  1. Match with the options

643\frac{64}{3}364​ corresponds to Option B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

PreviousNext

More from Area Under the Curves

  • The area of the region bounded by the curves x(1+y2)=1 and y2=2x is:2025 · MCQ
  • Let the area of the region (x,y):2y≤x2+3, y+∣x∣≤3, y≥∣x−1∣ be A. Then 6A is equal to :2025 · MCQ
  • Let the area enclosed between the curves ∣y∣=1−x2 and x2+y2=1 be α. If 9α=βπ+γ;β,γ are integers, then the value of ∣β−γ∣ equals:2025 · MCQ
  • The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :2024 · MCQ
  • Three points O(0,0),P(a,a2),Q(−b,b2),a>0, b>0, are on the parabola y=x2. Let S1​ be the area of the region…2024 · Numerical
  • The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y2=kx and ky2=2(y−x) is maximum, is equal to :2024 · Numerical
  • One of the points of intersection of the curves y=1+3x−2x2 and y=x1​ is (21​,2). Let the area of the region enclosed by these curves be 241​(l5​+m)−nloge​(1+5​)…2024 · MCQ
  • The area (in sq. units) of the region described by {(x,y):y2≤2x, and y≥4x−1} is2024 · MCQ