NOTE: Riemann integral

Riemann integral is a good start for learning integral as the first rigorous definition of integral on an interval.

Definition

Part 1

Let f(x)f(x)f(x) be a continuous function on the interval [a,b][a, b][a,b], we can make a mmm cut(m∈Nm \in \mathbb{N}m∈N) to get m+1m+1m+1 points on [a,b][a, b][a,b], they will be a series Δ=x0,x1,x2,x3,...,xm−1,xm\Delta = x_0, x_1, x_2, x_3, ..., x_{m-1}, x_mΔ=x0​,x1​,x2​,x3​,...,xm−1​,xm​. This series follows the following rule:

These points split [a,b][a, b][a,b] into mmm subintervals [xk−1,xk][x_{k-1}, x_k][xk−1​,xk​], where k=1,2,3,...,mk = 1, 2, 3, ..., mk=1,2,3,...,m. Now, forall kkk we pick a ξk\xi_kξk​ that xk−1≤ξk≤xkx_{k-1} \le \xi_k \le x_kxk−1​≤ξk​≤xk​, make a retangle with width is xk−xk−1x_k - x_{k-1}xk​−xk−1​ and height is f(ξk)f(\xi_k)f(ξk​). Thus, we can collect all retangles' area sum via the following formula:

By the max-min theorem, we can know there have MkM_kMk​ and μk\mu_kμk​ for f(x)f(x)f(x)'s max and min value on [xk−1,xk][x_{k-1}, x_k][xk−1​,xk​]. As above sum of retangle's area, but for MkM_kMk​ and μk\mu_kμk​(they are kind of ξk\xi_kξk​ of course), we have the following two formulas

• Maximum SΔS\DeltaSΔ

  SΔ=∑k=1mMk(xk−xk−1)S\Delta = \sum_{k=1}^{m} M_k(x_k-x_{k-1})SΔ=k=1∑m​Mk​(xk​−xk−1​)

• and Minimum sΔs\DeltasΔ

  sΔ=∑k=1mμk(xk−xk−1)s\Delta = \sum_{k=1}^{m} \mu_k(x_k-x_{k-1})sΔ=k=1∑m​μk​(xk​−xk−1​)

Since

, we know

Part 2

We know f(x)f(x)f(x) is uniformly continuous on [a,b][a, b][a,b]. So, for any ϵ∈R+\epsilon \in \mathbb{R}^+ϵ∈R+, exists a δ(ϵ)∈R+\delta(\epsilon) \in \mathbb{R}^+δ(ϵ)∈R+, let arbitrary two points xxx and x′x'x′, if the following formula is true

is true. And since

• μk=f(αk)\mu_k = f(\alpha_k)μk​=f(αk​)
• Mk=f(βk)M_k = f(\beta_k)Mk​=f(βk​)
• xk−1≤αk≤xkx_{k-1} \le \alpha_k \le x_kxk−1​≤αk​≤xk​
• xk−1≤βk≤xkx_{k-1} \le \beta_k \le x_kxk−1​≤βk​≤xk​

If xk−xk−1<δ(ϵ)x_k - x_{k-1} \lt \delta(\epsilon)xk​−xk−1​<δ(ϵ), then Mk−μk<ϵM_k - \mu_k \lt \epsilonMk​−μk​<ϵ. Thus, if we write the maximum value of interval's length as the follow(notice xk−xk−1x_k - x_{k-1}xk​−xk−1​ can be different for different kkk, since we didn't require same length for these subintervals):

Then once δ[Δ]<δ(ϵ)\delta[\Delta] \lt \delta(\epsilon)δ[Δ]<δ(ϵ), we have

Now, make a new ϵ′=ϵ/(b−a)\epsilon' = \epsilon/(b - a)ϵ′=ϵ/(b−a), we have δ(ϵ′)\delta(\epsilon')δ(ϵ′) that if δ[Δ]<δ(ϵ′)\delta[\Delta] \lt \delta(\epsilon')δ[Δ]<δ(ϵ′) then SΔ−sΔ<ϵ′S\Delta - s\Delta \lt \epsilon'SΔ−sΔ<ϵ′. Since ϵ′∈R\epsilon' \in \mathbb{R}ϵ′∈R, we can rewrite to δ[Δ]<δ(ϵ)\delta[\Delta] \lt \delta(\epsilon)δ[Δ]<δ(ϵ) then SΔ−sΔ<ϵS\Delta - s\Delta \lt \epsilonSΔ−sΔ<ϵ.

Part 3

We have a parition Δ\DeltaΔ, using the same way can get another parition Δ′\Delta'Δ′, their union is another parition Δ′′\Delta''Δ′′. Of course, there will also have σΔ\sigma\DeltaσΔ, σΔ′\sigma\Delta'σΔ′, and σΔ′′\sigma\Delta''σΔ′′ as sum of retangles from paritions.

Now, it's easy to find, Δ′′=x0′′,x1′′,x2′′,...,xp′′\Delta'' = x_0'', x_1'', x_2'', ..., x_p''Δ′′=x0′′​,x1′′​,x2′′​,...,xp′′​ can parition any [xk−1,xk][x_{k-1}, x_k][xk−1​,xk​] in Δ\DeltaΔ. Remember μk\mu_kμk​ is the minimum value of f(x)f(x)f(x) in [xk−1,xk][x_{k-1}, x_k][xk−1​,xk​], so μk≤f(ξp′′)\mu_k \le f(\xi_p'')μk​≤f(ξp′′​), where ξp′′∈Δ′′\xi_p'' \in \Delta''ξp′′​∈Δ′′. Then mapping xh′′=xk−1x_h'' = x_{k-1}xh′′​=xk−1​ and xj′′=xkx_j'' = x_kxj′′​=xk​. Thus, we get

We conclude

Now consider SΔ′S\Delta'SΔ′, we know σΔ′′≤SΔ′\sigma\Delta'' \le S\Delta'σΔ′′≤SΔ′ via same process, conclude

Let's consider all partitions Δ\DeltaΔ of [a,b][a, b][a,b], the corresponding sΔs\DeltasΔ of has a supremum

Obviously, s≤SΔ′s \le S\Delta's≤SΔ′, and since Δ′\Delta'Δ′ can be arbitrary parition so

Combine part 1 and part 2, conclude if δ[Δ]<δ(ϵ)\delta[\Delta] \lt \delta(\epsilon)δ[Δ]<δ(ϵ), then ∣σΔ−s∣<ϵ|\sigma\Delta - s| \lt \epsilon∣σΔ−s∣<ϵ.

Final

Now we can say for any partition, whatever how we pick ξk\xi_kξk​ for a partition Δ\DeltaΔ, we have

When δ[Δ]→0\delta[\Delta] \rightarrow 0δ[Δ]→0, say σ[Δ]=∑k=1mf(ξk)(xk−xk−1)\sigma[\Delta] = \sum_{k=1}^{m}f(\xi_k)(x_k - x_{k-1})σ[Δ]=∑k=1m​f(ξk​)(xk​−xk−1​) has limit sss, write as

We call it the definite integral of f(x)f(x)f(x) on [a,b][a, b][a,b], represent as

Vocabulary

1. f(x)f(x)f(x) is the integrand of ∫abf(x)dx\int_{a}^{b} f(x)dx∫ab​f(x)dx
2. ∫abf(x)dx\int_{a}^{b} f(x)dx∫ab​f(x)dx called the integrate of f(x)f(x)f(x) from aaa to bbb
3. aaa and bbb called the lower and upper limit of the definite integral