I assume that you are given sum_(i=1)^n i^5 = (n^2(2n^2+2n-1)(n+1)^2)/12n∑i=1i5=n2(2n2+2n−1)(n+1)212 and sum_(i=1)^n i^2 = (n(n+1)(2n+1))/6n∑i=1i2=n(n+1)(2n+1)6.
int_0^2 (x^5+x^2)dx∫20(x5+x2)dx
I'll use int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x
Delta x = (b-a)/n, in this case Delta x = 2/n
x_i = a+iDelta x, in this case x_i = i2/n.
Since f(x) = x^5+x^2, we have
int_0^2 (x^5+x^2)dx = lim_(nrarroo) sum_(i=1)^n (i^5 32/n^5 + i^2 4/n^2) 2/n
= lim_(nrarroo) (64/n^6sum_(i=1)^n i^5+ 8/n^3 sum_(i=1)^n i^2)
= lim_(nrarroo) (64/n^6 ( (n^2(2n^2+2n-1)(n+1)^2)/12) + 8/n^3 ((n(n+1)(2n+1))/6))
= lim_(nrarroo) (64/12((n^2(2n^2+2n-1)(n+1)^2)/n^6) + 8/6 ((n(n+1)(2n+1))/n^3))
=64/12(2)+8/6 (2) = 32/3+8/3 = 40/3
