- #1

Hummingbird25

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Hi All,

I have been given this example and am trying to explain.

[tex]\int_{0}^{1} {x^2} dx[/tex]

Solve this integral with respect to the definition of the defined integral.

(1)

Let [tex]P_{n}[/tex] be the partion of the interval [0,1] into n equally sized sub-partions.

Thus must mean that ?

Then [tex]P_n = \mathop {\lim }\limits_{n \to \infty } (\sum_{i=1}^{n} f(i/n)) \cdot 1/n[/tex]

since [tex]\triangle x = \frac{1-0}{n} = \frac{1}{n}[/tex]

2) Write the expression of the uppersum [tex]U(P_n)[/tex]

[tex]U(P_n) = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot (\sum_{i=1}^{n} (\frac{i}{n})^2] = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot [\frac{1}{n})^2 + (\frac{2}{n})^2 + \cdots + (\frac{n}{n})^2] = \mathop {\lim }\limits_{n \to +\infty } (\frac{1}{n^3} \cdot (1^2 + \cdots + n^2)) [/tex]

Then simplify the expressing using [tex] 1^2 + \cdots + n^2 = \frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}[/tex]

By insterting this into the final express for [tex]U(P_n)[/tex] I get [tex]U(P_n) = \mathop {\lim }\limits_{n \to +\infty } (\frac{1}{n^3} \cdot (\frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}) = \frac{1}{3} [/tex]. That must be the idear about simplifying the expression ;)

3) Write the Lower sum [tex]U(P_n)[/tex] and simply the expression using [tex] 1^2 + \cdots + n^2 = \frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}[/tex]

Isn't that simply

[tex]U(P_n) = [\mathop {\lim }\limits_{n \to -\infty } \frac{1}{n} \cdot (\sum_{i=1}^{n} (\frac{i}{n})^2] = \cdots = \mathop {\lim }\limits_{n \to -\infty } (\frac{1}{n^3} \cdot (\frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}) = \frac{1}{3}[/tex]

?

4)

Find the limit [tex]\mathop {\lim }\limits_{n \to \infty} U(\mathcal{P}_n)[/tex] and [tex]\mathop {\lim }\limits_{n \to \infty} L(\mathcal{P}_n)[/tex]

Since n tends to infinity in both the Lower and Upper Limit then [tex]\mathop {\lim }\limits_{n \to \infty} U(\mathcal{P}_n) = \mathop {\lim }\limits_{n \to \infty} L(\mathcal{P}_n)= \frac{1}{3}[/tex] ??

Finally show that the original integeral exists and write its value.

Assuming from the above that [tex]sup(L(\mathcal{P}_n) \leq imf(U(\mathcal{P}_n)[/tex] the the orginal integral

[tex]\int_{0}^{1} {x^2} dx[/tex] exists and more over the function [tex]x^2[/tex] is said to be Riemann integral over the interval [0,1]

and its value being written as [tex]\int_{0}^{1} {x^2} dx = sup(L(\mathcal{P}_n)) = imf(U(\mathcal{P}_n)) = \frac{1}{3} = \frac{1}{3} [/tex]

q.e.d.

Yours truely

Hummingbird

I have been given this example and am trying to explain.

[tex]\int_{0}^{1} {x^2} dx[/tex]

Solve this integral with respect to the definition of the defined integral.

## Homework Statement

(1)

Let [tex]P_{n}[/tex] be the partion of the interval [0,1] into n equally sized sub-partions.

Thus must mean that ?

## The Attempt at a Solution

Then [tex]P_n = \mathop {\lim }\limits_{n \to \infty } (\sum_{i=1}^{n} f(i/n)) \cdot 1/n[/tex]

since [tex]\triangle x = \frac{1-0}{n} = \frac{1}{n}[/tex]

## Homework Statement

2) Write the expression of the uppersum [tex]U(P_n)[/tex]

## The Attempt at a Solution

[tex]U(P_n) = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot (\sum_{i=1}^{n} (\frac{i}{n})^2] = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot [\frac{1}{n})^2 + (\frac{2}{n})^2 + \cdots + (\frac{n}{n})^2] = \mathop {\lim }\limits_{n \to +\infty } (\frac{1}{n^3} \cdot (1^2 + \cdots + n^2)) [/tex]

Then simplify the expressing using [tex] 1^2 + \cdots + n^2 = \frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}[/tex]

By insterting this into the final express for [tex]U(P_n)[/tex] I get [tex]U(P_n) = \mathop {\lim }\limits_{n \to +\infty } (\frac{1}{n^3} \cdot (\frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}) = \frac{1}{3} [/tex]. That must be the idear about simplifying the expression ;)

## Homework Statement

3) Write the Lower sum [tex]U(P_n)[/tex] and simply the expression using [tex] 1^2 + \cdots + n^2 = \frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}[/tex]

Isn't that simply

## The Attempt at a Solution

[tex]U(P_n) = [\mathop {\lim }\limits_{n \to -\infty } \frac{1}{n} \cdot (\sum_{i=1}^{n} (\frac{i}{n})^2] = \cdots = \mathop {\lim }\limits_{n \to -\infty } (\frac{1}{n^3} \cdot (\frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}) = \frac{1}{3}[/tex]

?

## Homework Statement

4)

Find the limit [tex]\mathop {\lim }\limits_{n \to \infty} U(\mathcal{P}_n)[/tex] and [tex]\mathop {\lim }\limits_{n \to \infty} L(\mathcal{P}_n)[/tex]

## The Attempt at a Solution

Since n tends to infinity in both the Lower and Upper Limit then [tex]\mathop {\lim }\limits_{n \to \infty} U(\mathcal{P}_n) = \mathop {\lim }\limits_{n \to \infty} L(\mathcal{P}_n)= \frac{1}{3}[/tex] ??

Finally show that the original integeral exists and write its value.

Assuming from the above that [tex]sup(L(\mathcal{P}_n) \leq imf(U(\mathcal{P}_n)[/tex] the the orginal integral

[tex]\int_{0}^{1} {x^2} dx[/tex] exists and more over the function [tex]x^2[/tex] is said to be Riemann integral over the interval [0,1]

and its value being written as [tex]\int_{0}^{1} {x^2} dx = sup(L(\mathcal{P}_n)) = imf(U(\mathcal{P}_n)) = \frac{1}{3} = \frac{1}{3} [/tex]

q.e.d.

Yours truely

Hummingbird

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