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A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx`is passing through origin. Then`g(1)=(3^7)/7`(b) `g(1)=(2^7)/7``g(-1)=1/7`(d) `g(-1)=(3^7)/(14)`A. `g(1)=(3^(7))/(7)`B. `g(1)=(2^(7))/(7)`C. `g(-1)=(1)/(7)`D. `g(-1)=(3^(7))/(14)` |
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Answer» Correct Answer - A::C `g(x)=int x^(27)(1+x+x^(2))^(6)(6x^(2)+5x+4)dx` `=int(x^(4)+x^(5)+x^(6))^(6)(6x^(5)+5x^(4)+4x^(3))dx ` `"Let " x^(6)+x^(5)+x^(4)=t " or "(6x^(5)+5x^(4)+4x^(3))dx=dt ` ` :. g(x)=int t^(6)dt=(t^(7))/(7)+C=(1)/(7)(x^(4)+x^(5)+x^(6))^(7)+C ` `g(0)=0 impliesC=0 impliesg(1)=(3^(7))/(7) " and " g(-1)=(1)/(7)` |
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