Write the Riemann sum to find the area under the graph of the function #f(x) = x^2# from #x = 1# to #x = 5#?

Question

Emily Ammerman · Accepted Answer

# int_1^5 \ x^2 \ dx = 124/3 #

We are asked to evaluate: # I = int_1^5 \ x^2 \ dx # Using Riemann sums. By definition of an integral, then # int_a^b \ f(x) \ dx # represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles. That is # int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)# And we partition the interval #[a,b]# equally spaced using: # Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } # # \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b-1 } # Here we have #f(x)=x^2# and so we partition the interval #[1,5]# using: # Delta = {1, 1+1(4)/n, 1+2 (4)/n, 1+3 (4)/n, ..., 4 } # And so: # I = int_1^5 \ (x^2) \ dx # # \ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ f(1+(4i)/n)# # \ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ (1+(4i)/n)^2# # \ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ ((n+4i)/n)^2# # \ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ 1/n^2(n+4i)^2# # \ \ = lim_(n rarr oo) 4/n^3 sum_(i=1)^n \ (n^2+8ni+16i^2)# # \ \ = lim_(n rarr oo) 4/n^3 {n^2 sum_(i=1)^n 1 + 8n sum_(i=1)^ni+16sum_(i=1)^ni^2)# Using the standard summation formula: # sum_(r=1)^n r \ = 1/2n(n+1) # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) # we have: # I = lim_(n rarr oo) 4/n^3 {n^3 + 8n 1/2n(n+1)+16 1/6n(n+1)(2n+1)# # \ \ = lim_(n rarr oo) 4/n^3 \ n/3 {3n^2 + 12n(n+1)+ 8(n+1)(2n+1)# # \ \ = lim_(n rarr oo) 4/(3n^2) (3n^2 + 12n^2+12n+ 16n^2+24n+8) # # \ \ = lim_(n rarr oo) 4/(3n^2) (31n^2+36n+ 8) # # \ \ = 4/3 lim_(n rarr oo) (31+36/n+ 8/n^2) # # \ \ = 4/3 (31+0+ 0) # # \ \ = 124/3 # Using Calculus If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get: # I = int_1^5 \ x^2 \ dx # # \ \ = [x^3/3]_1^5 # # \ \ = 125/3-1/3 # # \ \ = 124/3 #

Scarlett Allison · Answer

undefined The Riemann sum to find the area under the graph of the function $f(x) = x^2$ from $x = 1$ to $x = 5$ is given by:

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

where $\Delta x$ is the width of each subinterval and $x_i$ is a sample point in the $i$th subinterval.

In this case, the interval $[1, 5]$ is divided into $n$ subintervals of equal width:

$$\Delta x = \frac{5 - 1}{n} = \frac{4}{n}$$

The sample points $x_i$ are chosen to be the right endpoints of each subinterval. So, $x_i = 1 + i \Delta x$ for $i = 1, 2, ..., n$.

Substituting $f(x) = x^2$ into the Riemann sum formula, we get:

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