# Determine the value of expression (Calc-II)

1. May 19, 2014

### alingy1

I represented the expression with this integral :

integral of (1-x)^3 from 0 to 1.

I got 0, which makes no sense...

How should I proceed?

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2. May 19, 2014

### alingy1

Correction:

I represented this using:

sum of (1-i/n)^3(1/n), i=1..infinity.

I expand it, and apply summation rules.

Problem solved?

3. May 19, 2014

### HallsofIvy

It took me a moment to understand what you were doing! The problem you gave "Find the integral of (1- x)^3, from 0 to 1" can be done with a simple integration. But what you are doing is using the basic "Riemann sums" definition. Yes, if you divide the interval from 0 to 1 into n parts each part has length 1/n. And, at the right endpoint of the ith interval the function value is (1- i/n)^3 so the "area" of the thin rectangle there is (1- i/n)^3(1/n) and the entire area, and so the integral, is approximated by the sum.

No, i does not go from 1 to infinity- it goes from 1 to n and then, to get the actual integral, you take the limit as n goes to infinity.
You should have
$$\frac{1}{n}\sum_{i=1}^n \left( 1- 3\frac{i}{n}+ 3\frac{i^2}{n^2}- \frac{i^3}{n^3}\right)$$
$$\frac{1}{n}\sum_{i= 1}^n 1- \frac{3}{n^2}\sum_{i= 1}^n i+ \frac{3}{n^3}\sum_{i= 1}^n i^2-\frac{1}{n^4}\sum_{i= 1}^n i^3$$

4. May 19, 2014

### Saitama

I don't see how you get that. Do you see that it should be:
$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3$$

EDIT: Ok, your form is correct too. Now just convert it to a definite integral which is:
$$\int_0^1 (1-x)^3\,dx$$
There is no need to expand it.

Last edited: May 19, 2014
5. May 19, 2014

### alingy1

Hmm... If I evaluate the integral, I get 1/4. But that answer is just right if I assume n goes to infinity. If am asked to determine the value of this expression, should I not use hallsofivy's last expression and the summation formulas to expand the i's into n?

I think my final answer should contain n's.

Ultimately, the integral is irrelevant in this problem?

6. May 19, 2014

### SammyS

Staff Emeritus
Yes, use Halls's last expression.

Then use the summation formulas, the results of which will contain n, but not i .

7. May 19, 2014

### alingy1

What about the integral? Is it "useless" in this specific problem?

8. May 19, 2014

### SammyS

Staff Emeritus
I think Halls has the correct interpretation of what's expected. Use the basic "Riemann sums" definition of the integral - taking the limit as n → ∞ .

To check your answer, find the value of the following integral in the usual way, by finding the anti-derivative of the integrand, etc.
$\displaystyle \int_0^1 (1-x)^3\,dx$​

9. May 20, 2014

### alingy1

Why would it be take the infinite sum? The question is determine the value of this expression without saying as n goes to infinity.

10. May 20, 2014

### Saitama

We can do that too but this is easily done using definite integrals. If you insist, here's how. I will be using the form I shown before because that doesn't requires expanding the terms as HallsOfIvy has done.

The sum under consideration is:
$$\frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3=\frac{1}{n^4}\sum_{i=1}^{n-1}i^3$$
Since $\displaystyle \sum_{i=1}^{n-1}i^3=\left(\frac{(n-1)n}{2}\right)^2$, we have the following limit:
\begin{aligned} \lim_{n\rightarrow \infty} \frac{1}{n^4}\left(\frac{(n-1)n}{2}\right)^2 &= \lim_{n\rightarrow \infty} \frac{(n-1)^2}{4n^2}\\ &= \lim_{n\rightarrow \infty} \frac{1}{4}\left(1-\frac{1}{n}\right)^2 \\ &= \boxed{\dfrac{1}{4}}\\ \end{aligned}

11. May 20, 2014

### alingy1

I'm sorry. Perhaps I was not clear. My question is: the way you just did it supposes that n->infinity. But the original question is : determine the value of this expression in the picture. And in the picture, there is NO indication that n->infinity.
Do you understand what I'm trying to convey?

12. May 20, 2014

### Saitama

Ah ok, I thought this was another question related to Riemann sums. In that case, you leave it till the following step:
$$\frac{1}{4}\left(1-\frac{1}{n}\right)^2$$