How do you use the limit process to find the area of the region between the graph #y=x^2+1# and the x-axis over the interval [0,3]?

Question

Emily Ammerman · Accepted Answer

# int_0^3 \ x^2+1 \ dx = 12 #

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 } # Here we have #f(x)=x^2+1# and so we partition the interval #[0,3]# using: # Delta = {0, 3/n, 2. 3/n,3. 3/n, ..., 3 } # And so: # I = int_0^3 \ x^2+1 \ dx # # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(0+i*3/n)# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f((3i)/n)# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {((3i)/n)^2+1} # # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ { (9i^2)/n^2 + 1 } # # \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n \ (9i^2)/n^2 + sum_(i=1)^n \ 1 } # # \ \ = lim_(n rarr oo) 3/n {9/n^2sum_(i=1)^n \ i^2 + sum_(i=1)^n \ 1 } # Using the standard summation formula: # sum_(r=1)^n a \ = an # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) # we have: # I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1)+n } # # \ \ = lim_(n rarr oo) 3 {3/(2n^2) (2n^2+3n+1)+1 } # # \ \ = lim_(n rarr oo) 3/(2n^2) {3 (2n^2+3n+1)+2n^2 } # # \ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3+2n^2 } # # \ \ = lim_(n rarr oo) 3/(2n^2) {8n^2+9n+3 } # # \ \ = lim_(n rarr oo) (24n^2+27n+9 ) /(2n^2) # # \ \ = lim_(n rarr oo) (12+27/(2n) + 9/(2n^2)) # # \ \ = 12+0 + 0 # # \ \ = 12 # Using Calculus If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get: # int_0^3 \ x^2+1 \ dx = [ x^3/3 + x ]_0^3 # # " " = (9+3)-0 # # " " = 12#

William Abdalla · Answer

undefined To find the area of the region between the graph $ y = x^2 + 1 $ and the x-axis over the interval $[0,3]$ using the limit process, you can follow these steps:

1. Divide the interval $[0,3]$ into $n$ equal subintervals of width $ \Delta x = \frac{3}{n}$.
2. Choose sample points $x_i^*$ in each subinterval.
3. Calculate the area of each rectangle formed by a subinterval and the function values at the sample points.
4. Sum up the areas of all the rectangles to approximate the total area.
5. Take the limit as $n$ approaches infinity to find the exact area.

The formula for the area of each rectangle is $ A_i = f(x_i^*) \Delta x $, where $ f(x_i^*) $ represents the function value at the sample point $x_i^*$ within the $i$th subinterval.

The total area is given by the limit of the sum of the areas of all rectangles as $ n $ approaches infinity:

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

In this case, with $ f(x) = x^2 + 1 $, you substitute the values of $x_i^*$ into the function and sum up the areas of all rectangles. Then, take the limit as $n$ approaches infinity to find the exact area.