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Sagot :
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].
### 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].
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