# `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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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)`

`f(x(i))=2/(1+(i^5(1024))/n^5)+((4i)/n)/(1+(i^5(1024))/n^5)`

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`f(x(i))=(2n^5)/(n^5+1024i^5)+((4i)(n^4))/(n^5+1024i^5)`

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Thus,

`lim_(n->oo)sum_(i=1)^n((2n^5)/(n^5+1024i^5)+((4i)(n^4))/(n^5+1024i^5))(4/n)`

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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