Fill in the blank. Show your work on Exam 4 Required Work to be Shown.

Given: [tex]f(x)=1-x^2[/tex]

a. Find the formula for the Riemann sum obtained by dividing the interval [tex][0,1][/tex] into [tex]n[/tex] equal subintervals and using the right-hand endpoint for [tex]c_k[/tex]. (Write your answer on your paper and show all of your work. Simplify to two terms.)

b. Then find the limit of the sum as [tex]n \rightarrow \infty[/tex] to calculate the area under the curve.

The area under the curve is: [tex]\square[/tex]



Answer :

To solve this problem, we need to follow a series of steps to calculate the Riemann sum and then determine the limit as [tex]\( n \to \infty \)[/tex].

### Part (a): Riemann Sum Formula

Given the function [tex]\( f(x) = 1 - x^2 \)[/tex] on the interval [tex]\([0,1]\)[/tex]:

1. Divide the interval [tex]\([0,1]\)[/tex] into [tex]\( n \)[/tex] subintervals:
Each subinterval will have width [tex]\( \Delta x = \frac{1-0}{n} = \frac{1}{n} \)[/tex].

2. Right-hand endpoint for [tex]\( c_k \)[/tex]:
The right-hand endpoint for the [tex]\( k \)[/tex]-th subinterval is:
[tex]\[ x_k = a + k \Delta x \][/tex]
Since [tex]\( a = 0 \)[/tex], this becomes:
[tex]\[ x_k = 0 + k \cdot \frac{1}{n} = \frac{k}{n} \][/tex]

3. Formula for the Riemann sum:
The Riemann sum using right-hand endpoints is given by:
[tex]\[ S_n = \sum_{k=1}^{n} f(x_k) \Delta x \][/tex]
Plugging in the given [tex]\( f(x) \)[/tex] and [tex]\( x_k \)[/tex]:
[tex]\[ S_n = \sum_{k=1}^{n} f\left( \frac{k}{n} \right) \Delta x \][/tex]
[tex]\[ S_n = \sum_{k=1}^{n} \left( 1 - \left( \frac{k}{n} \right)^2 \right) \cdot \frac{1}{n} \][/tex]
[tex]\[ S_n = \sum_{k=1}^{n} \left( 1 - \frac{k^2}{n^2} \right) \cdot \frac{1}{n} \][/tex]
[tex]\[ S_n = \sum_{k=1}^{n} \left( \frac{1}{n} - \frac{k^2}{n^3} \right) \][/tex]
[tex]\[ S_n = \sum_{k=1}^{n} \frac{1}{n} - \sum_{k=1}^{n} \frac{k^2}{n^3} \][/tex]
[tex]\[ S_n = \frac{1}{n} \sum_{k=1}^{n} 1 - \frac{1}{n^3} \sum_{k=1}^{n} k^2 \][/tex]
[tex]\[ S_n = 1 - \frac{1}{n^3} \sum_{k=1}^{n} k^2 \][/tex]

Simplifying further, we need the formula for the sum of squares:
[tex]\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]

So the Riemann sum formula becomes:
[tex]\[ S_n = 1 - \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \][/tex]
[tex]\[ S_n = 1 - \frac{(n+1)(2n+1)}{6n^2} \][/tex]

### Part (b): Taking the Limit as [tex]\( n \to \infty \)[/tex]

Now, we need to find the limit of [tex]\( S_n \)[/tex] as [tex]\( n \to \infty \)[/tex]:

[tex]\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{(n+1)(2n+1)}{6n^2} \right) \][/tex]

Consider the limit:

[tex]\[ \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} \][/tex]
[tex]\[ = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} \][/tex]
[tex]\[ = \lim_{n \to \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} \][/tex]
[tex]\[ = \frac{2}{6} = \frac{1}{3} \][/tex]

Thus, we have:

[tex]\[ \lim_{n \to \infty} S_n = 1 - \frac{1}{3} = \frac{2}{3} \][/tex]

So, the area under the curve [tex]\( f(x) = 1 - x^2 \)[/tex] on the interval [tex]\([0,1]\)[/tex] is:

[tex]\[ \boxed{\frac{2}{3}} \][/tex]

Therefore, the final answer for the area under the curve is: [tex]\(\frac{2}{3}\)[/tex].