MHB Computing a Simple Integral Via Riemann Sums

[Math Amateur]

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
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Peter

 

[Ackbach]

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.

 

[Math Amateur]

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

 

[Euge]

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.

 

[Math Amateur]

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