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JEE Main 20172 April
Exam
JEE Main
Year
2017
Date
2 April
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2017 · 2 April · Mathematics · Area Under Curves

The area (in sq. units) of the region \ (x, y): x ≥ 0, x + y ≤ 3, x...

The area (in sq. units) of the region {(x,y):x≥0,x+y≤3,x2≤4 yandy ≤1+x}\left\{\left(x, y\right): x \ge 0, x + y \le 3, {x}^{2} \le 4\text{ yandy } \le 1 + \sqrt{x}\right\} isThe area (in sq. units) of the region \ (x, y): x ≥ 0, x + y ≤ 3, x ^ 2 ≤ 4 yandy ≤ 1 + x \ is

Options

  1. A5912sq.units\dfrac{59}{12}sq . units
  2. B32sq.units\dfrac{3}{2}sq . units
  3. C73sq.units\dfrac{7}{3}sq . units
  4. D52sq.units\dfrac{5}{2}sq . units

Correct answer

D. 52sq.units\dfrac{5}{2}sq . units

Step-by-step solution

The area (in sq. units) of the region \ (x, y): x ≥ 0, x + y ≤ 3, x ^ 2 ≤ 4 yandy ≤ 1 + x \ is

Required Area

=∫01(1+x)dx+∫12(3−x)dx−∫02x24dx= \mathop{∫}_{0}^{1}\left(1 + \sqrt{x}\right)dx + \mathop{∫}_{1}^{2}\left(3 - x\right)dx - \mathop{∫}_{0}^{2}\dfrac{{x}^{2}}{4}dx

=[x+23x32]01+[3x−x22]12−[x312]02= {\left[x + \dfrac{2}{3}x\dfrac{3}{2}\right]}_{0}^{1} + {\left[3x - \dfrac{{x}^{2}}{2}\right]}_{1}^{2} - {\left[\dfrac{{x}^{3}}{12}\right]}_{0}^{2}

=(1+23)+(4−52)−23= \left(1 + \dfrac{2}{3}\right) + \left(4 - \dfrac{5}{2}\right) - \dfrac{2}{3}

=52sq.units= \dfrac{5}{2}sq . units

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