How do you use limits to evaluate #int x^2dx# from [0,4]?

Question

Emma Aldridge · Accepted Answer

Please see below.

Here is a limit definition of the definite integral. (I hope it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks. .#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#. Where, for each positive integer #n#, we let #Deltax = (b-a)/n# And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.) We'll do one small step at a time. #int_0^4 x^2 dx#. Find #Delta x# For each #n#, we get #Deltax = (b-a)/n = (4-0)/n = 4/n# Find #x_i# And #x_i = a+iDeltax = 0+i4/n = (4i)/n# Find #f(x_i)# #f(x_i) = {:x_i:}^2 = ((4i)/n)^2# # = (16i^2)/n^2# Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sum. #sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ((16i^2)/n^2) 4/n# # = sum_(i=1)^n( (64i^2)/n^3)# # = 64/n^2 sum_(i=1)^n(i^2)# Evaluate the sum # = 64/n^3((n(n+1)(2n+1))/6)# (We used a summation formula in the previous step.) Rewrite before finding the limit #sum_(i=1)^n f(x_i)Deltax = 64/6((n(n+1)(2n+1))/n^3)# # = 32/3((n(n+1)(2n+1))/n^3)# Now we need to evaluate the limit as #nrarroo#. #lim_(nrarroo) ((n(n+1)(2n+1))/n^3)= lim_(nrarroo) (n/n (n+1)/n (2n+1)/n)# # = (1)(1)(2) = 2# To finish the calculation, we have #int_0^4 x^2dx = lim_(nrarroo)(32/3((n(n+1)(2n+1))/n^3))# # = 32/3(2) = 64/3#

Samantha Adkison · Answer

undefined To evaluate the integral ∫x^2 dx from 0 to 4 using limits, you can follow these steps:

1. Start with the indefinite integral: ∫x^2 dx.
2. Find the antiderivative of x^2, which is (1/3)x^3.
3. Evaluate the antiderivative at the upper limit of integration (4) and subtract the value when evaluated at the lower limit of integration (0).
4. Substitute the upper limit into the antiderivative: (1/3)(4)^3.
5. Substitute the lower limit into the antiderivative: (1/3)(0)^3.
6. Subtract the result of evaluating the antiderivative at the lower limit from the result of evaluating it at the upper limit.

So, the integral of x^2 from 0 to 4 equals:

(1/3)(4)^3 - (1/3)(0)^3
= (1/3)(64) - (1/3)(0)
= (64/3) - 0
= 64/3.

Therefore, the value of the integral ∫x^2 dx from 0 to 4 is 64/3.