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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Area Under Curves

The area (in sq. units) of the region \ (x, y) ∈ R ^ 2 |4 x ^ 2 ≤ y...

The area (in sq. units) of the region \left\{\left(x, y\right) \in {R}^{2}\left|4{x}^{2} \le y \le 8x + 12\right\right\} is

Options

  1. A1253\dfrac{125}{3}
  2. B1283\dfrac{128}{3}
  3. C1243\dfrac{124}{3}
  4. D1273\dfrac{127}{3}

Correct answer

B. 1283\dfrac{128}{3}

Step-by-step solution

4x2=y4{x}^{2} = y
y=8x+12y = 8x + 12
4x2=8x+124{x}^{2} = 8x + 12
x2−2x−3=0{x}^{2} - 2x - 3 = 0
x=−1,3x = - 1, 3
A=∫−13(8x+12−4x2)dxA = \mathop{∫}_{- 1}^{3}\left(8x + 12 - 4{x}^{2}\right)dx
A=8x22+12x−4x33A = \dfrac{8{x}^{2}}{2} + 12x - \dfrac{4{x}^{3}}{3}3−1\begin{array}{ll}3 \\ - 1\end{array}=(4(9)+36−36)−(4−12+43)=36+8−43= \left(4\left(9\right) + 36 - 36\right) - \left(4 - 12 + \dfrac{4}{3}\right) = 36 + 8 - \dfrac{4}{3}
=44−43=132−43=1283= 44 - \dfrac{4}{3} = \dfrac{132 - 4}{3} = \dfrac{128}{3}
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