Answer to Question #147650 in Calculus for liam donohue

Question

∫2bbx5dx =

Answer 1

We know that if $\intop f(x)dx=\phi(x)+C,$ then $\intop_p^qf(x)dx=\lbrack\phi(x)+C\rbrack_p^q$

$=\phi(q)-\phi(p)$ ....(1)

So according to the problem,we have

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

Now $\intop x^5dx= \lbrack x^6/6+C\rbrack$

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

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

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

$=(63/6)b^6$

$=(21/2)b^6$