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

2021 · 24 Feb · Shift 2 · Q33
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. /2021 · 24 Feb · Shift 2 · Q33

Area Under the Curves question

2021 · 24 Feb · Shift 2 · Q33

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the region : R={(x,y):5x2≤y≤2x2+9}R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\}R={(x,y):5x2≤y≤2x2+9} is :
  1. A
    636\sqrt 363​ square units
  2. B
    12312\sqrt 3123​ square units
  3. C
    11311\sqrt 3113​ square units
  4. D
    939\sqrt 393​ square units
View written solutionFree

Correct answer: B

  1. Identify the bounding curves

The region is R={(x,y):5x2≤y≤2x2+9}.R=\{(x,y): 5x^2 \le y \le 2x^2+9\}.R={(x,y):5x2≤y≤2x2+9}.

So the lower curve is y=5x2y=5x^2y=5x2 and the upper curve is y=2x2+9.y=2x^2+9.y=2x2+9.

The area exists where the upper curve is above the lower curve: 2x2+9≥5x2.2x^2+9 \ge 5x^2.2x2+9≥5x2.

  1. Find the points of intersection

Set the two curves equal: 5x2=2x2+95x^2=2x^2+95x2=2x2+9 3x2=93x^2=93x2=9 x2=3x^2=3x2=3 x=±3.x=\pm \sqrt{3}.x=±3​.

Thus the region is between x=−3x=-\sqrt{3}x=−3​ and x=3x=\sqrt{3}x=3​.

  1. Set up the area integral

Area between two curves is A=∫x1x2(upper−lower) dx.A=\int_{x_1}^{x_2} (\text{upper} - \text{lower})\,dx.A=∫x1​x2​​(upper−lower)dx.

Hence A=∫−33[(2x2+9)−5x2]dxA=\int_{-\sqrt{3}}^{\sqrt{3}} \big[(2x^2+9)-5x^2\big]dxA=∫−3​3​​[(2x2+9)−5x2]dx =∫−33(9−3x2) dx.=\int_{-\sqrt{3}}^{\sqrt{3}} (9-3x^2)\,dx.=∫−3​3​​(9−3x2)dx.

  1. Evaluate the integral

A=[9x−x3]−33.A=\left[9x-x^3\right]_{-\sqrt{3}}^{\sqrt{3}}.A=[9x−x3]−3​3​​.

Now,

  • At x=3x=\sqrt{3}x=3​: 93−(3)3=93−33=63.9\sqrt{3}-(\sqrt{3})^3=9\sqrt{3}-3\sqrt{3}=6\sqrt{3}.93​−(3​)3=93​−33​=63​.

  • At x=−3x=-\sqrt{3}x=−3​: 9(−3)−(−3)3=−93+33=−63.9(-\sqrt{3})-(-\sqrt{3})^3=-9\sqrt{3}+3\sqrt{3}=-6\sqrt{3}.9(−3​)−(−3​)3=−93​+33​=−63​.

Therefore, A=63−(−63)=123.A=6\sqrt{3}-(-6\sqrt{3})=12\sqrt{3}.A=63​−(−63​)=123​.

  1. Check options
  • A: 636\sqrt{3}63​ ❌
  • B: 12312\sqrt{3}123​ ✅
  • C: 11311\sqrt{3}113​ ❌
  • D: 939\sqrt{3}93​ ❌

So the correct option is B.

PreviousNext

More from Area Under the Curves

  • The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to ​.2021 · Numerical
  • The area (in sq. units) of the region, given by the set {(x,y)∈R×R∣x≥0,2x2≤y≤4−2x} is :2021 · MCQ
  • The area of the region S={(x,y):3x2≤4y≤6x+24} is ​.2021 · Numerical
  • Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 − 3x2 − 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ​…2021 · Numerical
  • The area bounded by the lines y = || x − 1 |− 2 | is ​.2021 · Numerical
  • Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = 2π​ in the first…2021 · MCQ
  • The area of the region bounded by the parabola (y − 2)2 = (x − 1), the tangent to it at the point whose ordinate is 3 and the x-axis is :2021 · MCQ
  • If the area of the bounded region R={(x,y):max{0,loge​x}≤y≤2x,21​≤x≤2} is , α(loge​2)−1+β(loge​2)+γ, then the value of (α+β−2λ)2…2021 · MCQ