An Analytical evaluation of a defined integral

In summary, Hummingbird is discussing the definition and properties of the Riemann integral with respect to the given example \int_{0}^{1} x^2 dx. They define the upper and lower sums and simplify them using the expression 1^2 + 2^2 + ... + n^2 = (n(n+1/2)(n+1))/3. They then find the limits of the upper and lower sums and show that the integral exists and has a value of 1/3.
  • #1
Hummingbird25
86
0
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.


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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  • #2
You seem to have exactly the same expressions for Lower sum and Upper sum and that is not right. What exactly is the difference between the Lower sum and Upper sum here?

For example, suppose you divide the interval from 0 to 1 into only 2 equal intervals. That is, your partition points are 0, 1/2, and 1. What are L(P2) and U(P[sub2[/sub])?
 
  • #3
Changed the result please look again!

First looking at the integral

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

1) Let [tex]\mathcal{P}_n[/tex] be the partion of the interval [0,1] into n-partions of equal size.

That means [tex]\mathcal{P}_n: \{0,1/n, 2/n,\ldots, (n-1)/n,1][/tex] which can the partioned into the sub-intervals

[tex][(i-1)/n, i/n][/tex], where [tex]i = 1,2, \ldots, n[/tex]

The Definition of the upper and lower sum says:

[tex]sup_{(x_{i-1},x_i)} f = M_i[/tex] and [tex]imf_{x_{i-1},x_i} f = m_i[/tex]

[tex]\mathcal{U}(f,\mathcal{P}_n}) = \sum_{i=1}^{n} M_i (x_{i} -x_{i-1})[/tex]

[tex]\mathcal{L}(f,\mathcal{P}_n}) = \sum_{i=1}^{n} m_i (x_{i} -x_{i-1})[/tex]

That leads me to question 2)

Write the [tex]\mathcal{U}(f,\mathcal{P}_n})[/tex] and simplify it using

[tex](1^2 + 2^2 + \ldots + n^2) = \frac{n(n+1/2)\cdot(n+1)}{3}[/tex]

First the upper sum [tex]\mathcal{U}(f,\mathcal{P}_n}) = \frac{1}{n} \cdot \sum_{i=1}^{n} (\frac{i}{n})^2 = \frac{1}{n^3} \cdot (1^2 + 2^2 + \ldots + n^2) = \frac{1}{2n} \ + \ \frac{1}{6n^2} \ + \ \frac{1}{3}[/tex] and the later being a more simplified version as required.

3) Write the Expression of the lower sum and simplify it like in the above.

[tex]\mathcal{L}(f,\mathcal{P}_n}) = \frac{1}{n} \cdot \sum_{i=1}^{n} (\frac{i-1}{n})^2 = \ldots = \frac{-1}{2n} \ + \ \frac{1}{6n^2} \ + \ \frac{1}{3}[/tex]

4)

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

By this
[tex]\mathop {\lim }\limits_{n \to \infty} \mathcal{L}(f,\mathcal{P}_n}) = \mathop {\lim }\limits_{n \to \infty} \frac{-1}{2n} \ + \ \frac{1}{6n^2} \ + \ \frac{1}{3} = \frac{1}{3}[/tex]

and

[tex]\mathop {\lim }\limits_{n \to \infty} \mathcal{U}(f,\mathcal{P}_n}) = \mathop {\lim }\limits_{n \to \infty} \frac{1}{2n} \ + \ \frac{1}{6n^2} \ + \ \frac{1}{3} = \frac{1}{3}[/tex]

Finally show that the integral exists and write its value!

Since the limits exists and that and that as shown above [tex]\mathcal{L}(f,\mathcal{P}_n}) \leq \mathcal{U}(f,\mathcal{P}_n})[/tex] the integral is said to be Riemann integrabel and exists.

The value of the integral is 1/3.

Have I covered what needs to be covered now Hall? Thanks in advanced.

Sincerely yOurs
Hummingbird...
 
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Related to An Analytical evaluation of a defined integral

1. What is a defined integral?

A defined integral is a type of mathematical calculation that involves finding the area under a curve or between two curves. It is represented by the symbol ∫ and is used to calculate the total accumulation of a function over a given interval.

2. How is a defined integral evaluated?

A defined integral is evaluated by first determining the limits of integration, which are the lower and upper bounds of the interval over which the integral is being calculated. Then, the function to be integrated is multiplied by an infinitesimal element called the differential, and the resulting expression is integrated using various techniques such as substitution, integration by parts, or the fundamental theorem of calculus.

3. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration and gives a numerical value as its result. On the other hand, an indefinite integral has no limits and gives a general expression as its result, which can be further evaluated using initial conditions or other methods.

4. Why is the concept of a defined integral important in mathematics?

The concept of a defined integral is important because it has numerous applications in various fields of science, technology, and engineering. It allows us to calculate important quantities such as area, volume, and work, and is also used in solving differential equations, optimization problems, and other real-world scenarios.

5. What are some common techniques used to evaluate defined integrals?

Some common techniques used to evaluate defined integrals include the substitution method, integration by parts, partial fractions, and the fundamental theorem of calculus. Each technique has its own advantages and is used depending on the complexity and nature of the integral.

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