Computing a Simple Integral Via Riemann Sums

In summary, the conversation discusses Exercise 2(a) from Manfred Stoll's book "Introduction to Real Analysis" and how to solve it using Riemann sums. The conversation includes a discussion on Stoll's definitions and notation, as well as a hint on how to solve the exercise by choosing a value of x in each subinterval. It is suggested to show that the given value of x is in the subinterval, which will simplify the solution. The conversation concludes with a helpful explanation and example of how to use the provided hint.
  • #1
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I am reading Manfred Stoll's book: Introduction to Real Analysis.

I need help with Exercise 2(a) from Stoll's Exercises 6.2 on page 229 ...

Exercise 2 reads as follows:
View attachment 3967
I was somewhat puzzled about how to do this exercise ... BUT ... even more puzzled when I read Stoll's hint for solving the exercise ... which reads as follows:
View attachment 3968
I would appreciate some help in framing a solution to this exercise ... and in particular getting some insight as to how Stoll means us to solve this given his "hint" ... ...

Hope someone can clarify this for me ... help will be appreciated ...
Stoll's definitions, notation and general approach to Riemann integration through Riemann sums is as follows:
View attachment 3969
View attachment 3970

Peter
 
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  • #2
So your interval is $[a,b]$, and you're integrating $x^2$. When you're doing Riemann sums, you can choose how you're going to pick an $x_i$ from each subinterval. Suppose you have $n$ equal-width subintervals of width
$$\Delta x=\frac{b-a}{n},$$
and you just pick the right-hand endpoint of each subinterval. Then $x_i=a+i \Delta x$, and your integral becomes
\begin{align*}
\int_a^b x^2 \, dx&=\lim_{n\to\infty}\sum_{i=1}^n (a+i\Delta x)^2 \, \Delta x \\
&=\lim_{n\to\infty}\sum_{i=1}^n \left(a+i\left(\frac{b-a}{n}\right)\right)^{\!2} \, \left(\frac{b-a}{n}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right)\sum_{i=1}^n \left(a^2+2ia\left(\frac{b-a}{n}\right)+i^2\left(\frac{b-a}{n}\right)^{\!2}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right) \left(\sum_{i=1}^na^2+\sum_{i=1}^n 2ia\left(\frac{b-a}{n}\right)+\sum_{i=1}^n i^2\left(\frac{b-a}{n}\right)^{\!2}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right) \left(a^2\sum_{i=1}^n 1+2a\left(\frac{b-a}{n}\right)\sum_{i=1}^n i +\left(\frac{b-a}{n}\right)^{\!2}\sum_{i=1}^n i^2\right) .
\end{align*}
These are standard sums now. Can you finish this part?

Now for the $x^n$ part, the hint there is all about choosing a value of $x$ in each subinterval. You need to show that the given value of $x$ is in the subinterval. Then, apparently, something will simplify and everything will come out great.
 
  • #3
Ackbach said:
So your interval is $[a,b]$, and you're integrating $x^2$. When you're doing Riemann sums, you can choose how you're going to pick an $x_i$ from each subinterval. Suppose you have $n$ equal-width subintervals of width
$$\Delta x=\frac{b-a}{n},$$
and you just pick the right-hand endpoint of each subinterval. Then $x_i=a+i \Delta x$, and your integral becomes
\begin{align*}
\int_a^b x^2 \, dx&=\lim_{n\to\infty}\sum_{i=1}^n (a+i\Delta x)^2 \, \Delta x \\
&=\lim_{n\to\infty}\sum_{i=1}^n \left(a+i\left(\frac{b-a}{n}\right)\right)^{\!2} \, \left(\frac{b-a}{n}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right)\sum_{i=1}^n \left(a^2+2ia\left(\frac{b-a}{n}\right)+i^2\left(\frac{b-a}{n}\right)^{\!2}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right) \left(\sum_{i=1}^na^2+\sum_{i=1}^n 2ia\left(\frac{b-a}{n}\right)+\sum_{i=1}^n i^2\left(\frac{b-a}{n}\right)^{\!2}\right) \\
&=\lim_{n\to\infty}\left(\frac{b-a}{n}\right) \left(a^2\sum_{i=1}^n 1+2a\left(\frac{b-a}{n}\right)\sum_{i=1}^n i +\left(\frac{b-a}{n}\right)^{\!2}\sum_{i=1}^n i^2\right) .
\end{align*}
These are standard sums now. Can you finish this part?

Now for the $x^n$ part, the hint there is all about choosing a value of $x$ in each subinterval. You need to show that the given value of $x$ is in the subinterval. Then, apparently, something will simplify and everything will come out great.
Thanks for the help Ackbach ...

just ts finished the first part ... Tedious but straightforward ...

Thanks again,

Peter
 
  • #4
Here's a way to use the hint. Let $f(x) = x^2$, $a\le x \le b$. Let $P : a = x_0 < x_1 < \cdots < x_n = b$ be an arbitrary partition of $[a,b]$. Then

$$x_{i-1}^2 = \frac{1}{3}(x_{i-1}^2 + x_{i-1}^2 + x_{i-1}^2) < \frac{1}{3}(x_{i-1}^2 + x_{i-1}x_i + x_i^2)$$ and similarly $$x_i^2 > \frac{1}{3}(x_{i-1}^2 + x_{i-1}x_i + x_i^2).$$

Thus

$$x_{i-1}^2 < t_i < x_i^2 \quad (1 \le i \le n).$$

Multiplying by $\Delta x_i$ then summing over all $1 \le i \le n$, we obtain

$$\sum_{i = 1}^n f(x_{i-1}) \Delta x_i < \sum_{i = 1}^n t_i \Delta x_i < \sum_{i = 1}^n f(x_i) \Delta x_i.$$

Since $t_i\Delta x_i = t_i(x_i - x_{i-1}) = x_i^3 - x_{i-1}^3$ for all $i$, the middle sum telescopes to $(x_n^3 - x_0^3)/3 = (b^2 - a^2)/3$. Thus

$$\sum_{i = 1}^n f(x_{i-1}) \Delta x_i < \frac{b^3 - a^3}{3} < \sum_{i = 1}^n f(x_i) \Delta x_i.$$

Since $P$ was an arbitrary partition of $[a,b]$, it follows that $\int_a^b x^2\, dx = (b^3 - a^3)/3$.
________________________

Sorry Peter, I didn't notice that you said you finished the first part.
 
  • #5
Euge said:
Here's a way to use the hint. Let $f(x) = x^2$, $a\le x \le b$. Let $P : a = x_0 < x_1 < \cdots < x_n = b$ be an arbitrary partition of $[a,b]$. Then

$$x_{i-1}^2 = \frac{1}{3}(x_{i-1}^2 + x_{i-1}^2 + x_{i-1}^2) < \frac{1}{3}(x_{i-1}^2 + x_{i-1}x_i + x_i^2)$$ and similarly $$x_i^2 > \frac{1}{3}(x_{i-1}^2 + x_{i-1}x_i + x_i^2).$$

Thus

$$x_{i-1}^2 < t_i < x_i^2 \quad (1 \le i \le n).$$

Multiplying by $\Delta x_i$ then summing over all $1 \le i \le n$, we obtain

$$\sum_{i = 1}^n f(x_{i-1}) \Delta x_i < \sum_{i = 1}^n t_i \Delta x_i < \sum_{i = 1}^n f(x_i) \Delta x_i.$$

Since $t_i\Delta x_i = t_i(x_i - x_{i-1}) = x_i^3 - x_{i-1}^3$ for all $i$, the middle sum telescopes to $(x_n^3 - x_0^3)/3 = (b^2 - a^2)/3$. Thus

$$\sum_{i = 1}^n f(x_{i-1}) \Delta x_i < \frac{b^3 - a^3}{3} < \sum_{i = 1}^n f(x_i) \Delta x_i.$$

Since $P$ was an arbitrary partition of $[a,b]$, it follows that $\int_a^b x^2\, dx = (b^3 - a^3)/3$.
________________________

Sorry Peter, I didn't notice that you said you finished the first part.
Thanks Euge ... That is MOST helpful ...

I did not know how to use the Hint provided ...

When I said that I had finished the first part, I meant that I had finished the calculations started so helpfully by Ackbach ...

Thanks again,

Peter
 

FAQ: Computing a Simple Integral Via Riemann Sums

What is a Riemann sum?

A Riemann sum is a method for approximating the area under a curve by dividing it into smaller rectangles and adding up the areas of those rectangles.

Why is computing a simple integral via Riemann sums important?

Computing a simple integral via Riemann sums is important because it allows us to approximate the area under a curve, which is a fundamental concept in calculus and has many real-world applications.

What is the difference between a left, right, and midpoint Riemann sum?

A left Riemann sum uses the left endpoint of each rectangle to calculate its height, a right Riemann sum uses the right endpoint, and a midpoint Riemann sum uses the midpoint. The choice of which type of Riemann sum to use can affect the accuracy of the approximation.

How do you calculate a Riemann sum?

To calculate a Riemann sum, you first divide the interval of integration into smaller subintervals. Then, for each subinterval, you choose a point (either left, right, or midpoint) and calculate the area of the rectangle formed by that point and the function. Finally, you add up all of the rectangle areas to get the approximation for the integral.

Are there any limitations to using Riemann sums to approximate integrals?

Yes, Riemann sums can only provide an approximation of the integral and may not be very accurate for functions with complex shapes. Additionally, as the number of subintervals increases, the calculations become more time-consuming and may not be feasible for very large intervals.

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