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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}

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66. Find the area lying above x-axis and included between the circle x² + y² = 8 x and inside of the parabola y² = 4 x.

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67. Calculate the area enclosed in the region:
$open curly brackets left parenthesis straight x comma space straight y right parenthesis space semicolon space space straight x squared plus straight y squared space less or equal than space 1 space less than space straight x plus 1 half straight y close curly brackets$

We are able to find the area included between the curves
$straight x squared plus straight y squared space equals space 1$ ...(1)
and $straight x plus 1 half straight y space equals space 1$ ...(2)
From (1) and (2), we get,
$open parentheses 1 minus 1 half straight y close parentheses squared space plus space straight y squared space equals space 1 space space space space space therefore space space space space 1 space plus straight y squared over 4 minus straight y plus straight y squared space equals space 1$
$therefore space space space 5 straight y squared minus 4 straight y space equals space 0 space space space space space rightwards double arrow space space space space space straight y left parenthesis 5 straight y minus 4 right parenthesis space equals space 0 space space rightwards double arrow space space space straight y space equals space 0 comma space 4 over 5 therefore space space from space left parenthesis 2 right parenthesis comma space space straight x space equals space 1 comma space space space 3 over 5$
$therefore$ circle (1) and line (2) intersect in points A (1, 0) and $straight B open parentheses 3 over 5 comma space 4 over 5 close parentheses$
Required area is shaded.
Required area = $integral subscript 3 divided by 5 end subscript superscript 1 square root of 1 minus straight x squared end root space dx minus integral subscript 3 divided by 5 end subscript superscript 1 2 space left parenthesis 1 minus straight x right parenthesis space dx$ $equals space open square brackets fraction numerator straight x square root of 1 minus straight x squared end root over denominator 2 end fraction plus 1 half sin to the power of negative 1 end exponent straight x close square brackets subscript 3 over 5 end subscript superscript 1 space plus space open square brackets left parenthesis 1 minus straight x squared right parenthesis close square brackets subscript 3 divided by 5 end subscript superscript 1 equals space open parentheses 0 plus 1 half sin to the power of negative 1 end exponent 1 close parentheses space minus space open parentheses 1 half 3 over 5 square root of 1 minus 9 over 25 end root plus 1 half sin to the power of negative 1 end exponent 3 over 5 close parentheses plus open parentheses 0 minus 4 over 25 close parentheses equals space 1 half. straight pi over 2 minus 3 over 10.4 over 5 minus straight i over 2 sin to the power of negative 1 end exponent 3 over 5 minus 4 over 25 equals space open parentheses straight pi over 4 minus 2 over 5 minus 1 half sin to the power of negative 1 end exponent 3 over 5 close parentheses space sq. space units. space$