How do you evaluate the definite integral by the limit definition given #int xdx# from [0,4]?

Question

Emilia Aguilar · 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.) I prefer to do this type of problem one small step at a time. #int_0^4 x 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 = (4i)/n# Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sums. #sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ( (4i)/n) 4/n# # = sum_(i=1)^n( (16i)/n^2)# # = 16/n^2 sum_(i=1)^n(i)# Evaluate the sums # = 16/n^2((n(n+1))/2)# (We used a summation formula for the sums in the previous step.) Rewrite before finding the limit #sum_(i=1)^n f(x_i)Deltax = 16/n^2((n(n+1))/2)# # = 8((n(n+1))/n^2))# Now we need to evaluate the limit as #nrarroo#. #lim_(nrarroo) ((n(n+1))/n^2) = lim_(nrarroo) (n/n*(n+1)/n) = 1# To finish the calculation, we have #int_0^4 x dx = lim_(nrarroo) ( 8((n(n+1))/n^2))# # = 8(1) = 8#

Ezekiel Alexander · Answer

undefined To evaluate the definite integral $ \int_0^4 x \, dx $ using the limit definition, we apply the following steps:

1. Divide the interval [0, 4] into subintervals.
2. Choose sample points within each subinterval.
3. Form a Riemann sum using these sample points.
4. Take the limit as the number of subintervals approaches infinity to find the definite integral.

For the interval [0, 4], we can choose any number of subintervals. Let's use n subintervals of equal width, denoted by Δx. The width of each subinterval is $ \Delta x = \frac{4 - 0}{n} = \frac{4}{n} $.

The sample points within each subinterval can be chosen arbitrarily, but for simplicity, we'll choose the right endpoint of each subinterval. So, the sample points will be $ x_i = i\Delta x $, where i ranges from 1 to n.

Now, we form the Riemann sum:

$$
R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} (i\Delta x) \Delta x
$$

This simplifies to:

$$
R_n = \sum_{i=1}^{n} i(\Delta x)^2
$$

Now, we take the limit as n approaches infinity:

$$
\lim_{n \to \infty} R_n = \lim_{n \to \infty} \sum_{i=1}^{n} i(\Delta x)^2
$$

Using the formula for the sum of the first n integers, $ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $, we have:

$$
\lim_{n \to \infty} R_n = \lim_{n \to \infty} \frac{n(n+1)}{2}(\Delta x)^2
$$

Substituting $ \Delta x = \frac{4}{n} $, we get:

$$
\lim_{n \to \infty} R_n = \lim_{n \to \infty} \frac{n(n+1)}{2} \left(\frac{4}{n}\right)^2
$$

Simplify and evaluate this limit to find the definite integral.