Using Riemann sums, evaluate [latex]\int_{0}^{1} {\left(x − x^{2}\right)} dx.[/latex]

Since the function [latex]f (x) = x − x^{2}[/latex] is continuous, it is integrable. Let [latex]P_{n} = \left\{x_{0}, x_{1}, x_{2}, ..., x_{n} \right\}[/latex] be the uniform partition of [0, 1], in which case [latex]x_{i} = \frac{i}{n}, i = 0, 1, . . . .n.[/latex] Choosing the mark [latex]α_{n} = (x_{0}, x_{1}, ..., x_{n−1})[/latex] ,we obtain [latex]S (f, P_{n}, α_{n}) = \sum\limits_{i=0}^{n−1}{\left(\frac{i}{n} − \frac{i^{2}}{n^{2}}\right)} \frac{1}{n}[/latex] [latex]= \frac{1}{n^{2}} \sum\limits_{i=0}^{n−1}{i} − \frac{1}{n^{3}} \sum\limits_{i=0}^{n−1}{i^{2}}[/latex] [latex]= \frac{1}{n^{2}} \frac{n (n − 1)}{2} − \frac{1}{n^{3}} \frac{(n − 1) n (2n − 1)}{6},[/latex] and since [latex]\left\|P_{n}\right\| = 1/n → 0,[/latex] we have [latex]\int_{0}^{1} {\left(x − x^{2}\right)} dx = \underset{n→∞}{\lim} S (f, P_{n}, α_{n})[/latex] [latex]= \frac{1}{2} − \frac{1}{3} = \frac{1}{6}.[/latex]

Question 8.5: Using Riemann sums, evaluate ∫0^1 (x − x²) dx.

Elements of Real Analysis

Using Riemann sums, evaluate $\int_{0}^{1} {\left(x − x^{2}\right)} dx.$

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Since the function $f (x) = x − x^{2}$ is continuous, it is integrable. Let $P_{n} = \left\{x_{0}, x_{1}, x_{2}, …, x_{n} \right\}$ be the uniform partition of [0, 1], in which case

$x_{i} = \frac{i}{n}, i = 0, 1, . . . .n.$

Choosing the mark $α_{n} = (x_{0}, x_{1}, …, x_{n−1})$ ,we obtain

$S (f, P_{n}, α_{n}) = \sum\limits_{i=0}^{n−1}{\left(\frac{i}{n} − \frac{i^{2}}{n^{2}}\right)} \frac{1}{n}$

$= \frac{1}{n^{2}} \sum\limits_{i=0}^{n−1}{i} − \frac{1}{n^{3}} \sum\limits_{i=0}^{n−1}{i^{2}}$

$= \frac{1}{n^{2}} \frac{n (n − 1)}{2} − \frac{1}{n^{3}} \frac{(n − 1) n (2n − 1)}{6},$

and since $\left\|P_{n}\right\| = 1/n → 0,$ we have

$\int_{0}^{1} {\left(x − x^{2}\right)} dx = \underset{n→∞}{\lim} S (f, P_{n}, α_{n})$

$= \frac{1}{2} − \frac{1}{3} = \frac{1}{6}.$