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Example 28 - Chapter 7 Class 12 Integrals
Last updated at December 16, 2024 by Teachoo
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Example 1 (i)
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Chapter 7 Class 12 Integrals
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Example 28 Evaluate ā«_(ā1)^2ā|š„^3āš„| šš„ |š„^3āš„|=|š„(š„^2ā1)| =|š„| |š„^2ā1| =|š„| |(š„ā1)(š„+1)| =|š„| |š„ā1||š„+1| Thus, š„=0, š„=1 & š„=ā1 ā“ |š„^3āš„|= {ā((āš„)(ā(š„ā1))(ā(š„+1)) šš š„<ā1@(āš„)"(ā(š„ā1))" (š„+1) šš ā1ā¤š„<0@(š„)"(ā(š„ā1))" (š„+1) šš 0ā¤š„<1@(š„)(š„ā1)(š„+1) šš š„ā„1)ā¤ |š„^3āš„|= {ā(āš„(š„ā1)(š„+1) šš š„<ā1@š„(š„ā1)(š„+1) šš ā1ā¤š„<0@āš„(š„ā1)(š„+1) šš 0ā¤š„<1@š„(š„ā1)(š„+1) šš š„ā„1)ā¤ |š„^3āš„|= {ā(ā(š„^3āš„) šš š„<ā1@(š„^3āš„) šš ā1ā¤š„<0@ā(š„^3āš„) šš 0ā¤š„<1@(š„^3āš„) šš š„ā„1)ā¤ Thus, our integration becomes ā«_(ā1)^2ā|š„^3āš„| šš„ =ā«_(ā1)^0ā(š„^3āš„) šš„āā«_0^1ā(š„^3āš„) šš„+ā«_1^2ā(š„^3āš„) šš„ =ā«_(ā1)^0āš„^3 šš„āā«_(ā1)^0āš„ šš„āā«_0^1āš„^3 šš„+ā«_0^1āš„ šš„+ā«_1^2āš„^3 šš„āā«_1^2āš„ šš„ =[š„^4/4]_(ā1)^0ā[š„^2/2]_(ā1)^0ā[š„^4/4]_0^1+[š„^2/2]_0^1+[š„^4/4]_1^2ā[š„^2/2]_1^2 =[(0 ā (ā1)^4)/4]ā[(0 ā (ā1)^2)/2]ā[((1)^4 ā 0)/4]+[((1)^2 ā 0)/2]+[((2)^4 ā (1)^4)/4]ā[((2)^2 ā (1)^2)/2] =(ā1)/4 ā [(ā1)/2] ā 1/4 + 1/2 + [(16 ā 1)/4]ā[(4 ā 1)/2] =(ā1)/4 + 1/2 ā 1/4 + 1/2 +15/4 ā 3/2 =(ā1 + 2 ā 1 + 2 + 15 ā 6)/4 =šš/š
