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QJEEM20Ma3c7f: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3c7f Solution: The area (in sq. units) of the region { (x, y) : 0 y x 2 + 1, 0 …

The area (in sq. units) of the region

{ (x, y) : 0 ≤\le y ≤\le x2 + 1, 0 ≤\le y ≤\le x + 1,

12{1 \over 2} ≤\le x ≤\le 2 } is :

Multiple Choice Options

  • A 7916{{79} \over {16}}
  • B 7924{{79} \over {24}}
  • C 236{{23} \over {6}}
  • D 2316{{23} \over {16}}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Area Under The Curves Question 101 English Explanation A=∫121(x2+1)dx+A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + } ∫12\int\limits_1^2 {} (x+1)dx{(x + 1)dx}

= [x33+x]121+[x22+x]12{\left[ {{{{x^3}} \over 3} + x} \right]_{{1 \over 2}}^1 + \left[ {{{{x^2}} \over 2} + x} \right]_1^2}

=(43−1324)+(4−32){ = \left( {{4 \over 3} - {{13} \over {24}}} \right) + \left( {4 - {3 \over 2}} \right)}

=1924+52{ = {{19} \over {24}} + {5 \over 2}}

=7924{ = {{79} \over {24}}}

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