Using Upper, Lower Sum, Prove the Following

[Astrum]

Homework Statement

$$f(x)=x^{2}$$ from [a,b]

Prove that $$F(x)=\frac{b^{3}}{2}-\frac{a^{3}}{2}$$

Homework Equations

The Attempt at a Solution

Using the definition of an integral, we get:
$$U(f,P)=\sum^{n}_{i=1}M_{i}(t_{i}-t_{i-1})$$
$$L(f,P)\sum^{n}_{i=1}m_{i}(t_{i}-t_{i-1})$$

for the function x², how do we find M?

I'm somewhat confused, a nudge in the right directions is welcomed.

 

[CompuChip]

Divide the interval [a,b] into n pieces and let t be the boundaries of these pieces; so t₀ = a, t₁ = a + (b - a) / n, t₂ = a + 2 (b - a) / n, etc.
M_i is the maximum of f on the piece [t_i-1, t_i] and m_i is the minimum.

It may become clearer if you draw a picture in which you draw the graph, the division of [a, b] and the rectangular areas that make up U(f, P) or L(f, P)

 

[Astrum]

Divide the interval [a,b] into n pieces and let t be the boundaries of these pieces; so t₀ = a, t₁ = a + (b - a) / n, t₂ = a + 2 (b - a) / n, etc.
M_i is the maximum of f on the piece [t_i-1, t_i] and m_i is the minimum.

It may become clearer if you draw a picture in which you draw the graph, the division of [a, b] and the rectangular areas that make up U(f, P) or L(f, P)

Ah, that's right, the partition is a=t₀<t₁<...<t_n=b

So we get this:
$$\sum^{n}_{i=1}M_{i}(a+\frac{b-a}{n})$$

But I still am not understanding where we get M and m from. We need to find the maximum minimum points on the interval? Or is there an easier way?

 

[HallsofIvy]

As long as a and b are positive, $x^2$ is an increasing function so the maximum value on each interval, $M_i$, is at the right end and the minimum value on each interval, $m_i$, is at the left end. Further, if you divide [a, b] into n intervals, each interval has length $t_i- t_{i-1}= (b- a)/n$, the left endpoint is a+ (b- a)i/n for i= 0 to n-1, and the right endpoint is a+ (b- a)i/n for i= 1 to n. That is, $M_i= (a+ (b-a)i/n)^2$, for i= 1 to n and $m_i= (a+ (b-a)i/n)^2$ for i= 0 to n-1.
$M_i(t_i- t_{i-1})= ((b-a)/n)(a+ (b-a)i/n)^2$ for i= 1 to n and $m_i(t_i- t_{i-1})= ((b- a)/n)(a+ (b- a)i/n)^2$
Multiply those out and sum. It will help if you know that $\sum_{i=1}^n c= nc$, $\sum_{i=1}^n i= n(n+ 1)/2$ and $\sum_{i=1}^n i^2= [n(n+1)(2n+1)]/6$. Those should be in your text.

 

[Astrum]

As long as a and b are positive, $x^2$ is an increasing function so the maximum value on each interval, $M_i$, is at the right end and the minimum value on each interval, $m_i$, is at the left end. Further, if you divide [a, b] into n intervals, each interval has length $t_i- t_{i-1}= (b- a)/n$, the left endpoint is a+ (b- a)i/n for i= 0 to n-1, and the right endpoint is a+ (b- a)i/n for i= 1 to n. That is, $M_i= (a+ (b-a)i/n)^2$, for i= 1 to n and $m_i= (a+ (b-a)i/n)^2$ for i= 0 to n-1.
$M_i(t_i- t_{i-1})= ((b-a)/n)(a+ (b-a)i/n)^2$ for i= 1 to n and $m_i(t_i- t_{i-1})= ((b- a)/n)(a+ (b- a)i/n)^2$
Multiply those out and sum. It will help if you know that $\sum_{i=1}^n c= nc$, $\sum_{i=1}^n i= n(n+ 1)/2$ and $\sum_{i=1}^n i^2= [n(n+1)(2n+1)]/6$. Those should be in your text.

You had some formatting problems XD

I think I get what you're saying though, it makes sense. So, in the interval [a,b], the highest value of f(x) will be at b. If I understand what you're saying, M will be equal to t_i², and m will be t²_i-1?

There seems to be some weird latex problem.

 

[Astrum]

I think I got it.

$$U(f,P)=\sum^{n}_{i=1}M_{i}(t_{i}-t_{i-1})$$
$$L(f,P)=\sum^{n}_{i=1}m{i-1}(t_{i}-t_{i-1})$$

$m_{i}=t^{2}_{i-1}; M_{i}=t^{2}_{i}$

$$=\sum^{n}_{i=1}t^{2}_{i}(\frac{b-a}{n})$$
$$=\sum^{n}_{i=1}t^{2}_{i-1}(\frac{b-a}{n})$$

$$t^{2}_{i}=i^{2}\frac{b^{2}-a^{2}}{n^{2}}$$
$$t^{2}_{i-1}=(i-1)^{2}\frac{b^{2}-a^{2}}{n^{2}}$$

$$=\sum^{n}_{i=1}i^{2}\frac{b^{2}-a^{2}}{n^{2}}(\frac{b-a}{n})$$
$$=\sum^{n}_{i=1}(i-1)^{2}\frac{b^{2}-a^{2}}{n^{2}}(\frac{b-a}{n})$$

Now we just need to simplify, right?