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# I always hv a confusion in solving func. Ques ? What is the basic concept to be applied ? One such ques...Let p(x) be a func. defined on R such that p`(x)=p`(1-x),for all x€(0,1),p(0)=1andp(1)=41.Then integration of p(x)dx from 1 to 0 equals to.....Plzz give me the basic idea

vishal
38 Points
3 years ago
P`(x)=[P(b)-P(a)]/(b-a)P`(x)=[P(1)-P(0)]/(1-0) =(41-1)/1=40Integrate w r to x- P(x)=40x+cAt x=0 or x=1 P(0)=0+c1=cThen, P(x)=40x+1Integrate w r to x from 0 to 1 =40x^2/2+x =(40/2+1)-(0+0) =20+1 =21
Manan Jain
29 Points
3 years ago
P`(x)=[P(b)-P(a)]/(b-a)P`(x)=[P(1)-P(0)]/(1-0) =(41-1)/1=40Integrate w r to x- P(x)=40x+cAt x=0 or x=1 P(0)=0+c1=cThen, P(x)=40x+1Integrate w r to x from 0 to 1 =40x^2/2+x =(40/2+1)-(0+0) =20+1 =21