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Application of Integrals

Name: Zigya - For The Curious Learner
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Short Answer Type

61.

Find the area of the region {(x, y): x² + y² ≤ 1 ≤ x + y}.

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Long Answer Type

62.

Find the area of the region bounded by the circle x² + y² = 1 and x + y = 1. Also draw a rough sketch.

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63.

Find the area of the region {(x, y): x² ≤ y ≤ |x|}.
Or
Find the area of the region bounded by the parabola y = x² and y = |x|.

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64. Draw the rough sketch and find the area of the region:
{(x, y) : y² ≤ 8 x, x² + x² ≤ 9}

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65. Find the area of the region {(x, y): y² ≤ 4 x, 4x² + 4 y² ≤ 9}

Given region is
   $open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y squared space less or equal than space 4 space straight x comma space space 4 straight x squared plus 4 straight y squared less or equal than 9 close curly brackets$
Consider the equations
$straight y squared space equals 4 space straight x$ ...(1)
and    $4 straight x squared plus 4 straight y squared equals space 9$
i.e., $straight x squared plus straight y squared space equals 9 over 4$ ...(2)
From (1) and (2), we get,
   $straight x squared plus 4 straight x equals 9 over 4 space or space 4 straight x squared plus 16 straight x minus 9 space equals space 0$
$therefore space space space space straight x space equals fraction numerator negative 16 plus-or-minus square root of 256 plus 144 end root over denominator 8 end fraction equals fraction numerator negative 16 plus-or-minus 20 over denominator 8 end fraction equals 1 half comma space minus 9 over 2$
$therefore space space from space left parenthesis 1 right parenthesis comma space space space space space straight y squared space equals 1 half space space space space space space space space space space space space space space space space space open square brackets because straight y squared equals negative 9 over 2 space does space not space give space real space points close square brackets because space straight y space equals space fraction numerator 1 over denominator square root of 2 end fraction comma space minus fraction numerator 1 over denominator square root of 2 end fraction$
$therefore$ curve (1) and (2) intersect in the points
   $straight P open parentheses 1 half comma space fraction numerator 1 over denominator square root of 2 end fraction close parentheses space space space and space space space straight Q open parentheses 1 half comma space minus fraction numerator 1 over denominator square root of 2 end fraction close parentheses$
From P, draw $PM space perpendicular space straight x minus axis$
Here OA = $3 over 2$

Required area = Area of shaded region
= 2 (area of region OAPO)
= 2 [area of region OMPO + area of region MAPM]
   $equals space 2 open square brackets integral subscript 0 superscript 1 divided by 2 end superscript 2 square root of straight x space dx plus integral subscript 1 divided by 2 end subscript superscript 3 divided by 2 end superscript square root of 9 over 4 minus straight x squared end root close square brackets space equals space 4 integral subscript 0 superscript 1 divided by 2 end superscript straight x to the power of 1 divided by 2 end exponent plus 2 space integral subscript 1 divided by 2 end subscript superscript 3 divided by 2 end superscript square root of open parentheses 3 over 2 close parentheses squared minus straight x squared end root dx$
   $equals 4 open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript 1 half end superscript plus 2 open square brackets fraction numerator straight x square root of begin display style 9 over 4 end style minus straight x squared end root over denominator 2 end fraction plus open parentheses begin display style 3 over 2 end style close parentheses squared over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 3 divided by 2 end fraction close parentheses close square brackets subscript 1 half end subscript superscript 3 over 2 end superscript$
$equals space 8 over 3 open square brackets straight x to the power of 3 divided by 2 end exponent close square brackets subscript 0 superscript 1 divided by 2 end superscript plus 2 open square brackets 1 half straight x square root of 9 over 4 minus straight x squared end root plus 9 over 8 sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 3 end fraction close parentheses close square brackets subscript 1 half end subscript superscript 3 over 2 end superscript equals space 8 over 3 open square brackets open parentheses 1 half close parentheses to the power of 3 over 2 end exponent minus 0 close square brackets space plus 2 open square brackets open parentheses 0 plus 9 over 8 sin to the power of negative 1 end exponent 1 close parentheses minus open curly brackets 1 fourth square root of 9 over 4 minus 1 fourth end root plus 9 over 8 sin to the power of negative 1 end exponent 1 third close curly brackets close square brackets equals space 8 over 3 fraction numerator 1 over denominator 2 square root of 2 end fraction plus 2 open square brackets 9 over 8 cross times straight pi over 2 minus fraction numerator 1 over denominator 2 square root of 2 end fraction minus 9 over 8 sin to the power of negative 1 end exponent open parentheses 1 third close parentheses close square brackets equals space fraction numerator 4 over denominator 3 square root of 2 end fraction plus fraction numerator 9 straight pi over denominator 8 end fraction minus fraction numerator 1 over denominator square root of 2 end fraction minus 9 over 4 sin to the power of negative 1 end exponent open parentheses 1 third close parentheses equals space open parentheses fraction numerator 4 square root of 2 over denominator 6 end fraction minus fraction numerator square root of 2 over denominator 2 end fraction close parentheses plus fraction numerator 9 straight pi over denominator 8 end fraction minus 9 over 4 space sin to the power of negative 1 end exponent open parentheses 1 third close parentheses space equals fraction numerator square root of 2 over denominator 6 end fraction plus fraction numerator 9 straight pi over denominator 8 end fraction minus 9 over 4 sin to the power of negative 1 end exponent open parentheses 1 third close parentheses$