Answer to Question #147650 in Calculus for liam donohue

Question #147650
∫2bbx5dx =
1
Expert's answer
2020-12-08T16:27:28-0500

We know that if f(x)dx=ϕ(x)+C,\intop f(x)dx=\phi(x)+C, then pqf(x)dx=[ϕ(x)+C]pq\intop_p^qf(x)dx=\lbrack\phi(x)+C\rbrack_p^q

=ϕ(q)ϕ(p)=\phi(q)-\phi(p) ....(1)

So according to the problem,we have

f(x)=x5,p=b,q=2bf(x)=x^5, p=b,q=2b

Now x5dx=[x6/6+C]\intop x^5dx= \lbrack x^6/6+C\rbrack

Therefore , b2b(x5)dx=[x6/6]b2b\int _b^2b (x^5) dx =[x^6/6]_b^2b

=(2b)6/6(b6)/6=(2b)^6/6-(b^6)/6

=(1/6)[64b6b6]=(1/6)[64b^6-b^6]

=(63/6)b6=(63/6)b^6

=(21/2)b6=(21/2)b^6


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