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JEE Main 202225 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2022 · 25 July · Morning Shift · Mathematics · Area Under Curves

The area of the region given by A = \ (x, y): x ^ 2 ≤ y ≤ min\ x +...

The area of the region given by A={(x,y):x2≤y≤min{x+2,4−3x}}A = \left\{\left(x, y\right): {x}^{2} \le y \le min\left\{x + 2, 4 - 3x\right\}\right\} isThe area of the region given by A = \ (x, y): x ^ 2 ≤ y ≤ min\ x + 2, 4 - 3x\ \ is

Options

  1. A318\dfrac{31}{8}
  2. B176\dfrac{17}{6}
  3. C196\dfrac{19}{6}
  4. D278\dfrac{27}{8}

Correct answer

B. 176\dfrac{17}{6}

Step-by-step solution

The graph of y=x2 and y=min{x+2,4−3x}y = {x}^{2} \text{ and } y = min\left\{x + 2, 4 - 3x\right\} will be

The area of the region given by A = \ (x, y): x ^ 2 ≤ y ≤ min\ x + 2, 4 - 3x\ \ is

From the graph,

required area A=∫−112(x+2−x2)dx+∫121(4−3x−x2)dxA = {∫}_{- 1}^{\dfrac{1}{2}}\left(x + 2 - {x}^{2}\right)dx + {∫}_{\dfrac{1}{2}}^{1}\left(4 - 3x - {x}^{2}\right)dx

=(x22+2x−x33)−112+(4x−3x22−x33)121=176= {\left(\dfrac{{x}^{2}}{2} + 2x - \dfrac{{x}^{3}}{3}\right)}_{- 1}^{\dfrac{1}{2}} + {\left(4x - 3\dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3}\right)}_{\dfrac{1}{2}}^{1} = \dfrac{17}{6}

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