Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 30

int_1^10 (x-4ln(x))dx
To express this definite integral as limit of Riemann's Sum, apply the formula:
int_a^b f(x) dx = lim_(n-> oo)sum_(i=1)^oo f(x_i)Delta x
where
Delta x = (b-a)/n
x_i = a + iDeltax
For the given definite interval, the Delta x is:
Delta x= (10-1)/n
Delta x=9/n
And its x_i is:
x_i= 1 + i *9/n
x_i=1+(9i)/n
x_i=(n+9i)/n
Plug-in this x_i to the function.
The function in the given integral is:
f(x) = x-4lnx
So, f(x_i) is:
f(x_i) = (n+9i)/n-4ln((n+9i)/n)
So the given definite integral
int_1^10 (x-4lnx) dx
becomes:
= lim_(n->oo) sum_(i=1)^n ((n+9i)/n -4ln((n+9i)/n)) * 9/n
=lim_(n->oo) sum_(i=1)^n (9(n+9i))/n^2-36/nln((n+9i)/n)

Therefore, int_1^10 (x-4lnx) dx=lim_(n-gtoo) sum_(i=1)^n (9(n+9i))/n^2-36/nln((n+9i)/n) .