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CBSE Class 12-science Answered

evaluate
 
integral subscript 1 superscript 4 left curly bracket vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 4 vertical line right curly bracket d x
 
Asked by ppratim02 | 20 Nov, 2016, 11:12: AM
answered-by-expert Expert Answer
begin mathsize 16px style Let space straight I equals integral subscript 1 superscript 4 open parentheses open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar plus open vertical bar straight x minus 4 close vertical bar close parentheses space dx Now comma space let space straight f left parenthesis straight x right parenthesis equals open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar plus open vertical bar straight x minus 4 close vertical bar Here comma space we space have space three space critical space points space that space is comma space straight x space equals space 1 comma space 2 comma space 4. To space remove space the space modulus space sign comma space we space consider space the space following space cases colon left parenthesis straight i right parenthesis space When space straight x less than 1 left parenthesis ii right parenthesis space When space 1 less or equal than straight x less than 2 left parenthesis iii right parenthesis space When space 2 less or equal than straight x less than 4 left parenthesis ii right parenthesis space When space straight x greater or equal than 4 So comma straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative open parentheses straight x minus 1 close parentheses minus open parentheses straight x minus 2 right parenthesis minus left parenthesis straight x minus 4 close parentheses comma space if space straight x less than 1 end cell row cell open parentheses straight x minus 1 close parentheses minus open parentheses straight x minus 2 right parenthesis minus left parenthesis straight x minus 4 close parentheses comma space if space 1 less or equal than straight x less than 2 end cell row cell open parentheses straight x minus 1 close parentheses plus open parentheses straight x minus 2 right parenthesis minus left parenthesis straight x minus 4 close parentheses comma space if space 2 less or equal than straight x less than 4 end cell row cell open parentheses straight x minus 1 close parentheses plus open parentheses straight x minus 2 right parenthesis plus left parenthesis straight x minus 4 close parentheses comma space if space straight x greater or equal than 4 end cell end table close rightwards double arrow straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative 3 straight x plus 7 comma space if space straight x less than 1 end cell row cell negative straight x plus 5 comma space if space 1 less or equal than straight x less than 2 end cell row cell straight x plus 1 comma space if space 2 less or equal than straight x less than 4 end cell row cell 3 straight x minus 7 comma space if space straight x greater or equal than 4 end cell end table close Thus comma space straight I equals integral subscript 1 superscript 4 open parentheses open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar plus open vertical bar straight x minus 4 close vertical bar close parentheses space dx rightwards double arrow straight I equals integral subscript 1 superscript 4 straight f left parenthesis straight x right parenthesis space dx rightwards double arrow straight I equals integral subscript 1 superscript 2 straight f left parenthesis straight x right parenthesis space dx plus integral subscript 2 superscript 4 straight f left parenthesis straight x right parenthesis space dx rightwards double arrow straight I equals integral subscript 1 superscript 2 left parenthesis negative straight x plus 5 right parenthesis space dx plus integral subscript 2 superscript 4 left parenthesis straight x plus 1 right parenthesis space dx rightwards double arrow straight I equals open square brackets 5 straight x minus straight x squared over 2 close square brackets squared subscript 1 plus open square brackets straight x squared over 2 plus straight x close square brackets to the power of 4 subscript 2 rightwards double arrow straight I equals open square brackets open parentheses 10 minus 4 over 2 close parentheses minus open parentheses 5 minus 1 half close parentheses close square brackets plus open square brackets open parentheses 16 over 2 plus 4 close parentheses minus open parentheses 4 over 2 plus 2 close parentheses close square brackets rightwards double arrow straight I equals open square brackets open parentheses 10 minus 2 close parentheses minus open parentheses 9 over 2 close parentheses close square brackets plus open square brackets open parentheses 8 plus 4 close parentheses minus open parentheses 2 plus 2 close parentheses close square brackets rightwards double arrow straight I equals open square brackets 7 over 2 close square brackets plus open square brackets 8 close square brackets rightwards double arrow straight I equals 23 over 2 end style
Answered by Rebecca Fernandes | 20 Nov, 2016, 09:05: PM

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