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Vector Algebra

Question
CBSEENMA12035682

Sketch the region bounded by the curves straight y equals square root of 5 minus straight x squared end root space and space straight y space equals open vertical bar straight x minus 1 close vertical bar and find its area using integration. 

Solution

Consider the given equation. 
straight y equals square root of 5 minus straight x squared end root
This equation represents a semicircle with centre at the origin and radius  = square root of 5 space units
Given that the region is bounded by the above semicircle and the line straight y equals open vertical bar straight x minus 1 close vertical bar
Let us find the point of intersection of the given curve meets the line straight y equals open vertical bar straight x minus 1 close vertical bar
rightwards double arrow square root of 5 minus straight x squared end root space equals space open vertical bar straight x minus 1 close vertical bar
Squaring both the sides, we have,
5 minus straight x squared space equals open vertical bar straight x minus 1 close vertical bar squared
rightwards double arrow 5 minus straight x squared space equals space straight x squared plus 1 minus 2 straight x
rightwards double arrow 2 straight x squared minus 2 straight x minus 5 plus 1 space equals space 0
rightwards double arrow 2 straight x squared minus 2 straight x minus 4 space equals space 0
rightwards double arrow straight x squared minus straight x minus 2 space equals 0
rightwards double arrow space straight x squared minus 2 straight x plus straight x minus 2 space equals 0
rightwards double arrow straight x left parenthesis straight x minus 2 right parenthesis plus 1 left parenthesis straight x minus 2 right parenthesis equals 0
rightwards double arrow left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x minus 2 right parenthesis space equals space 0
rightwards double arrow straight x space equals negative 1 comma space straight x space equals space 2
When space straight x space equals space minus 1 comma space space straight y space equals space 2
When space straight x space equals space 2 comma space space straight y space equals space 1
Consider the following figure
Thus the intersection points are 1,2 and 2,1 ( ) ( ) Consider the following sketch of the bounded region.

Required Area, straight A space equals space integral subscript negative 1 end subscript superscript 2 left parenthesis straight y subscript 2 minus straight y subscript 1 right parenthesis dx space
equals integral subscript negative 1 end subscript superscript 1 space open square brackets square root of 5 minus straight x squared end root plus left parenthesis straight x minus 1 right parenthesis close square brackets dx space plus space integral subscript 1 superscript 2 space open square brackets square root of 5 minus straight x squared end root minus left parenthesis straight x minus 1 right parenthesis close square brackets dx
equals integral subscript negative 1 end subscript superscript 1 space square root of 5 minus straight x squared end root dx plus integral subscript negative 1 end subscript superscript 1 xdx space minus integral subscript negative 1 end subscript superscript 1 dx space plus space integral subscript 1 superscript 2 square root of 5 minus straight x squared end root dx minus integral subscript 1 superscript 2 xdx plus integral subscript 1 superscript 2 dx
equals open square brackets straight x over 2 square root of 5 minus straight x squared end root plus 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 5 end fraction close parentheses close square brackets subscript negative 1 end subscript superscript 1 space plus space open parentheses straight x squared over 2 close parentheses subscript negative 1 end subscript superscript 1 space minus left parenthesis straight x right parenthesis subscript negative 1 end subscript superscript 1 plus open square brackets straight x over 2 square root of 5 minus straight x squared end root plus 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 5 end fraction close parentheses close square brackets subscript 1 superscript 2
minus open parentheses straight x squared over 2 close parentheses subscript 1 superscript 2 space plus left parenthesis straight x right parenthesis subscript 1 superscript 2
equals 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 5 end fraction close parentheses plus 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator square root of 5 end fraction close parentheses minus 1 half
Required space Area space equals open square brackets 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 5 end fraction close parentheses plus 5 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator 2 over denominator square root of 5 end fraction close parentheses minus 1 half close square brackets sq. space units

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