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JEE Main 20199 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2019 · 9 January · Evening Shift · Mathematics · Area Under Curves

The area of the region A = (x, y): 0 ≤ y ≤ x|x| + 1and - 1 ≤ x ≤ 1...

The area of the region A=(x,y):0≤y≤x∣x∣+1and−1≤x≤1A = { \left(x, y\right): 0 \le y \le x\left|x\right| + 1and - 1 \le x \le 1 } in sq. units, isThe area of the region A = (x, y): 0 ≤ y ≤ x|x| + 1and - 1 ≤ x ≤ 1 in sq. units, is

Options

  1. A43\dfrac{4}{3}
  2. B22
  3. C13\dfrac{1}{3}
  4. D23\dfrac{2}{3}

Correct answer

B. 22

Step-by-step solution

x\left|x\right| = \left\{x2,x≥0−x2,x<0\begin{array}{ll}{x}^{2}, x \ge 0 \\ - {x}^{2}, x < 0\end{array}\right)

The area of the region A = (x, y): 0 ≤ y ≤ x|x| + 1and - 1 ≤ x ≤ 1 in sq. units, is

Area enclosed =∫−10(−x2+1)dx+∫01(x2+1)dx= \mathop{∫}_{- 1}^{0}\left(- {x}^{2} + 1\right)dx + \mathop{∫}_{0}^{1}\left({x}^{2} + 1\right)dx

=[−x33+x]−10+[x33+x]01= {\left[- \dfrac{{x}^{3}}{3} + x\right]}_{- 1}^{0} + {\left[\dfrac{{x}^{3}}{3} + x\right]}_{0}^{1}

=2= 2

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