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using integration, find the area bounded by the curves y=|x-1| and y=3-|x|. 

Asked by Karishma 19th March 2015, 1:59 AM
Answered by Expert
Answer:
C o n s i d e r space t h e space c u r v e s space y equals open vertical bar x minus 1 close vertical bar space a n d space y equals 3 minus open vertical bar x close vertical bar 3 minus open vertical bar x close vertical bar equals open vertical bar x minus 1 close vertical bar  W h e n space x less than 0 rightwards double arrow 3 minus open parentheses negative x close parentheses equals open parentheses negative x plus 1 close parentheses rightwards double arrow 3 plus x equals negative x plus 1 rightwards double arrow x plus x equals plus 1 minus 3 rightwards double arrow 2 x equals negative 2 rightwards double arrow x equals negative 1 S u b s t i t u t e space x equals negative 1 space i n space y equals open vertical bar x minus 1 close vertical bar comma space w e space h a v e comma space y equals open vertical bar negative 1 minus 1 close vertical bar equals open vertical bar negative 2 close vertical bar equals 2  W h e n space x greater than 0 rightwards double arrow 3 minus x equals x minus 1 rightwards double arrow 3 plus 1 equals 2 x rightwards double arrow 2 x equals 4 rightwards double arrow x equals 2 S u b s t i t u t e space x equals 2 space i n space y equals open vertical bar x minus 1 close vertical bar comma space w e space h a v e comma space y equals open vertical bar 2 minus 1 close vertical bar equals 1  T h u s space t h e space p o i n t s space o f space i n t e r s e c t i o n space a r e space open parentheses negative 1 comma 2 close parentheses space a n d space open parentheses 2 comma 1 close parentheses  R e q u i r e d space a r e a equals integral subscript negative 1 end subscript superscript 0 open parentheses 3 plus x plus x minus 1 close parentheses d x plus integral subscript 0 superscript 1 open parentheses 3 minus x plus x minus 1 close parentheses d x plus integral subscript 1 superscript 2 open parentheses 3 minus x minus x plus 1 close parentheses d x equals integral subscript negative 1 end subscript superscript 0 open parentheses 2 x plus 2 close parentheses d x plus integral subscript 0 superscript 1 2 d x plus integral subscript 1 superscript 2 open parentheses 4 minus 2 x close parentheses d x equals open parentheses fraction numerator 2 x squared over denominator 2 end fraction close parentheses subscript negative 1 end subscript superscript 0 plus open parentheses 2 x close parentheses subscript negative 1 end subscript superscript 0 plus open parentheses 2 x close parentheses subscript 0 superscript 1 plus open parentheses 4 x close parentheses subscript 1 superscript 2 minus open parentheses fraction numerator 2 x squared over denominator 2 end fraction close parentheses subscript 1 superscript 2 equals negative 1 plus 2 open parentheses negative open parentheses negative 1 close parentheses close parentheses plus 2 plus 4 open parentheses 2 minus 1 close parentheses minus open parentheses 2 squared minus 1 squared close parentheses equals negative 1 plus 2 plus 2 plus 4 minus open parentheses 4 minus 1 close parentheses equals negative 1 plus 2 plus 2 plus 4 minus 3 equals 4 space s q. space u n i t s
Answered by Expert 20th March 2015, 1:03 PM
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