`int_2^6x/(1 + x^5)dx` Express the integral as a limit of Riemann sums. Do not evaluate the limit.

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mitch1 eNotes educator| Certified Educator

You have to recall the definition of the Reiman Integral `int_a^bf(x)dx=lim_(n->oo)sum_(i=1)^nf(x(i))Deltax`

`where Deltax =(b-a)/n and x(i)= a +iDeltax`

`x `

`a=2 and b = 6 `

`Deltax = (6-2)/n= 4/n`

`x(i) = 2 + i4/n`

`f(x(i))= (2+i(4/n))/(1+(i(4/n))^5)`

Simplifying it will be 

`f(x(i)) = 2/(1+i^5(4/n)^5)+i(4/n)/(1+i^5(4/n)^5)`


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mitch1 eNotes educator| Certified Educator

Sorry for the slip, it should be 


`lim_(n->oo)sum_(i=1)^n(2/(1+(2+(4i)/n)^5) +((4i)/n)/(1+(2+(4i/n))^5))((4i)/n)`

I forgot to type  2 + 4i/n  for  the x  in the bottom of the  fraction .

Please disregard the simplifying part . Take note that you will just plug-in 

the x(i)  that you get in the function  then simplify after that