How do you use limits to evaluate #int(x^5+x^2)dx# from [0,2]?

1 Answer
Jan 6, 2017

Answer:

Please see the explanation section below.

Explanation:

I assume that you are given #sum_(i=1)^n i^5 = (n^2(2n^2+2n-1)(n+1)^2)/12# and #sum_(i=1)^n i^2 = (n(n+1)(2n+1))/6#.

#int_0^2 (x^5+x^2)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#