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

Question

Emily Ammerman · Accepted Answer

Please see the explanation section below.

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#

Emma Aldridge · Answer

undefined To evaluate the integral ∫(x^5 + x^2) dx from 0 to 2 using limits, you can follow these steps:

1. Find the antiderivative of the function, which is (1/6)x^6 + (1/3)x^3.
2. Evaluate the antiderivative at the upper limit (2) and subtract the value obtained by evaluating it at the lower limit (0).
3. Plug in the upper limit (2) into the antiderivative: (1/6)(2^6) + (1/3)(2^3).
4. Plug in the lower limit (0) into the antiderivative: (1/6)(0^6) + (1/3)(0^3).
5. Subtract the result obtained in step 4 from the result obtained in step 3.

After calculation, you will find that the integral evaluates to (1/6)(2^6) + (1/3)(2^3) - (1/6)(0^6) - (1/3)(0^3) = (64/3) units.