We know that if ∫f(x)dx=ϕ(x)+C,\intop f(x)dx=\phi(x)+C,∫f(x)dx=ϕ(x)+C, then ∫pqf(x)dx=[ϕ(x)+C]pq\intop_p^qf(x)dx=\lbrack\phi(x)+C\rbrack_p^q∫pqf(x)dx=[ϕ(x)+C]pq
=ϕ(q)−ϕ(p)=\phi(q)-\phi(p)=ϕ(q)−ϕ(p) ....(1)
So according to the problem,we have
f(x)=x5,p=b,q=2bf(x)=x^5, p=b,q=2bf(x)=x5,p=b,q=2b
Now ∫x5dx=[x6/6+C]\intop x^5dx= \lbrack x^6/6+C\rbrack∫x5dx=[x6/6+C]
Therefore , ∫b2b(x5)dx=[x6/6]b2b\int _b^2b (x^5) dx =[x^6/6]_b^2b∫b2b(x5)dx=[x6/6]b2b
=(2b)6/6−(b6)/6=(2b)^6/6-(b^6)/6=(2b)6/6−(b6)/6
=(1/6)[64b6−b6]=(1/6)[64b^6-b^6]=(1/6)[64b6−b6]
=(63/6)b6=(63/6)b^6=(63/6)b6
=(21/2)b6=(21/2)b^6=(21/2)b6
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