# An Analytical evaluation of a defined integral

1. Feb 11, 2008

### Hummingbird25

Hi All,

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

$$\int_{0}^{1} {x^2} dx$$

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

1. The problem statement, all variables and given/known data
(1)

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

Thus must mean that ?

3. The attempt at a solution

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

since $$\triangle x = \frac{1-0}{n} = \frac{1}{n}$$

1. The problem statement, all variables and given/known data
2) Write the expression of the uppersum $$U(P_n)$$

3. The attempt at a solution
$$U(P_n) = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot (\sum_{i=1}^{n} (\frac{i}{n})^2] = [\mathop {\lim }\limits_{n \to +\infty } \frac{1}{n} \cdot [\frac{1}{n})^2 + (\frac{2}{n})^2 + \cdots + (\frac{n}{n})^2] = \mathop {\lim }\limits_{n \to +\infty } (\frac{1}{n^3} \cdot (1^2 + \cdots + n^2))$$

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

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

1. The problem statement, all variables and given/known data
3) Write the Lower sum $$U(P_n)$$ and simply the expression using $$1^2 + \cdots + n^2 = \frac{n \cdot (n+\frac{1}{2})\cdot(n+1)}{3}$$

Isn't that simply

3. The attempt at a solution

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

?
1. The problem statement, all variables and given/known data
4)

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

3. The attempt at a solution
Since n tends to infinity in both the Lower and Upper Limit then $$\mathop {\lim }\limits_{n \to \infty} U(\mathcal{P}_n) = \mathop {\lim }\limits_{n \to \infty} L(\mathcal{P}_n)= \frac{1}{3}$$ ??

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

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

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

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

q.e.d.

Yours truely
Hummingbird

Last edited: Feb 11, 2008
2. Feb 11, 2008

### HallsofIvy

Staff Emeritus
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. Feb 11, 2008

### Hummingbird25

Changed the result please look again!

First looking at the integral

$$\int_{0}^{1} x^2 dx$$

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

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

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

The Definition of the upper and lower sum says:

$$sup_{(x_{i-1},x_i)} f = M_i$$ and $$imf_{x_{i-1},x_i} f = m_i$$

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

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

That leads me to question 2)

Write the $$\mathcal{U}(f,\mathcal{P}_n})$$ and simplify it using

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

First the upper sum $$\mathcal{U}(f,\mathcal{P}_n}) = \frac{1}{n} \cdot \sum_{i=1}^{n} (\frac{i}{n})^2 = \frac{1}{n^3} \cdot (1^2 + 2^2 + \ldots + n^2) = \frac{1}{2n} \ + \ \frac{1}{6n^2} \ + \ \frac{1}{3}$$ 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.

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

4)

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

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

and

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

Finally show that the integral exists and write its value!

Since the limits exists and that and that as shown above $$\mathcal{L}(f,\mathcal{P}_n}) \leq \mathcal{U}(f,\mathcal{P}_n})$$ 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....

Last edited: Feb 11, 2008