MathIsimple
Course 11

Definite Integration

Section 1: Riemann Integral

Definition 1.1: Partition

A partition Δ\Delta of [a,b][a, b] is a finite set of points:

a=x0<x1<x2<⋯<xn=ba = x_0 < x_1 < x_2 < \cdots < x_n = b

The subintervals are [xk−1,xk][x_{k-1}, x_k] with lengths Δxk=xk−xk−1\Delta x_k = x_k - x_{k-1}.

Definition 1.2: Riemann Sum

For a partition Δ\Delta and sample points ξk∈[xk−1,xk]\xi_k \in [x_{k-1}, x_k], the Riemann sum is:

SΔ(f,ξ)=∑k=1nf(ξk)ΔxkS_\Delta(f, \xi) = \sum_{k=1}^{n} f(\xi_k) \Delta x_k
Definition 1.3: Riemann Integral

A function ff is Riemann integrable on [a,b][a, b] if there exists a number II such that:

∀ε>0,∃δ>0:if ∥Δ∥<δ, then ∣SΔ(f,ξ)−I∣<ε\forall \varepsilon > 0, \exists \delta > 0: \text{if } \|\Delta\| < \delta, \text{ then } |S_\Delta(f, \xi) - I| < \varepsilon

We write ∫abf(x) dx=I\int_a^b f(x)\,dx = I.

Theorem 1.1: Integrability of Continuous Functions

If ff is continuous on [a,b][a, b], then ff is Riemann integrable on [a,b][a, b].

Section 2: Properties of Definite Integrals

Theorem 2.1: Linearity
∫ab[f(x)+g(x)] dx=∫abf(x) dx+∫abg(x) dx\int_a^b [f(x) + g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx
∫abkf(x) dx=k∫abf(x) dx\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx
Theorem 2.2: Additivity
∫abf(x) dx+∫bcf(x) dx=∫acf(x) dx\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx
Theorem 2.3: Reversal
∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx
Example 2.1: Basic Computation

Evaluate: ∫01x2 dx\int_0^1 x^2\,dx

Solution: Using FTC with F(x)=x3/3F(x) = x^3/3:

∫01x2 dx=x33∣01=13−0=13\int_0^1 x^2\,dx = \frac{x^3}{3}\bigg|_0^1 = \frac{1}{3} - 0 = \frac{1}{3}

Section 3: Fundamental Theorem of Calculus

Theorem 3.1: FTC Part 1

If ff is continuous on [a,b][a, b], then the function:

F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt

is differentiable on (a,b)(a, b) and:

F′(x)=f(x)F'(x) = f(x)
Proof of FTC Part 1:

For h>0h > 0:

F(x+h)−F(x)h=1h∫xx+hf(t) dt\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt

By MVT for integrals, ∃c∈[x,x+h]\exists c \in [x, x+h] with:

∫xx+hf(t) dt=f(c)⋅h\int_x^{x+h} f(t)\,dt = f(c) \cdot h

As h→0h \to 0, c→xc \to x, so:

F′(x)=lim⁡h→0f(c)hh=f(x)F'(x) = \lim_{h \to 0} \frac{f(c)h}{h} = f(x)
∎
Theorem 3.2: FTC Part 2 (Newton-Leibniz Formula)

If FF is an antiderivative of ff on [a,b][a, b], then:

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)
Proof of Newton-Leibniz Formula:

Let G(x)=∫axf(t) dtG(x) = \int_a^x f(t)\,dt. By FTC Part 1, G′(x)=f(x)G'(x) = f(x).

Since F′(x)=f(x)F'(x) = f(x) and G′(x)=f(x)G'(x) = f(x), we have (F−G)′(x)=0(F - G)'(x) = 0.

Therefore F(x)−G(x)=CF(x) - G(x) = C for some constant CC.

Since G(a)=0G(a) = 0, we have F(a)=CF(a) = C.

Thus F(b)−G(b)=F(a)F(b) - G(b) = F(a), so G(b)=F(b)−F(a)G(b) = F(b) - F(a). ∎

∎
Example 3.1: FTC Application

Evaluate: ∫0πsin⁡x dx\int_0^\pi \sin x\,dx

Solution: Since (−cos⁡x)′=sin⁡x(-\cos x)' = \sin x:

∫0πsin⁡x dx=−cos⁡x∣0π=−cos⁡π+cos⁡0=1+1=2\int_0^\pi \sin x\,dx = -\cos x\big|_0^\pi = -\cos \pi + \cos 0 = 1 + 1 = 2

Section 4: Mean Value Theorem for Integrals

Theorem 4.1: Mean Value Theorem for Integrals

If ff is continuous on [a,b][a,b], then there exists c∈[a,b]c \in [a,b] such that:

∫abf(x) dx=f(c)(b−a)\int_a^b f(x)\,dx = f(c)(b-a)
Example 4.1: Application

For f(x)=x2f(x) = x^2 on [0,2][0,2]:

∫02x2 dx=83=f(c)⋅2\int_0^2 x^2\,dx = \frac{8}{3} = f(c) \cdot 2

So c2=43c^2 = \frac{4}{3}, giving c=233c = \frac{2\sqrt{3}}{3}.

Section 5: Integration by Substitution (Definite)

Theorem 5.1: Substitution Rule for Definite Integrals

If u=g(x)u = g(x) and the derivative of gg is continuous on [a,b][a,b], then:

∫abf(g(x))g′(x) dx=∫g(a)g(b)f(u) du\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du
Example 5.1: Substitution

Evaluate: ∫01x(1+x2)3 dx\int_0^1 x(1+x^2)^3\,dx

Let u=1+x2u = 1+x^2, then du=2x dxdu = 2x\,dx:

∫12u32 du=18(16−1)=158\int_1^2 \frac{u^3}{2}\,du = \frac{1}{8}(16-1) = \frac{15}{8}

Section 6: Integration by Parts (Definite)

Theorem 6.1: Integration by Parts for Definite Integrals
∫abu dv=uv∣ab−∫abv du\int_a^b u\,dv = uv\big|_a^b - \int_a^b v\,du
Example 6.1: Integration by Parts

Evaluate: ∫0πxsin⁡x dx\int_0^\pi x\sin x\,dx

Let u=xu = x, dv=sin⁡x dxdv = \sin x\,dx:

=−xcos⁡x∣0π+∫0πcos⁡x dx=π+sin⁡x∣0π=π= -x\cos x\big|_0^\pi + \int_0^\pi \cos x\,dx = \pi + \sin x\big|_0^\pi = \pi

Section 7: Additional Properties

Theorem 7.1: Additivity

For any c∈[a,b]c \in [a,b]:

∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx
Theorem 7.2: Comparison

If f(x)≤g(x)f(x) \leq g(x) on [a,b][a,b], then:

∫abf(x) dx≤∫abg(x) dx\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx

Work These Ideas by Hand — Then Check Yourself

Practice Quiz: Definite Integration
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1
What is a partition of [a,b][a,b]?
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The Riemann sum SΔ(f,ξ)S_\Delta(f, \xi) equals:
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A bounded function ff is Riemann integrable on [a,b][a,b] iff:
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The Newton-Leibniz formula states:
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If ff is continuous at x0x_0, then F′(x0)F'(x_0) equals:
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ddx∫ax2sin⁡t dt\frac{d}{dx}\int_a^{x^2} \sin t\,dt equals:
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A continuous function on [a,b][a,b] is:
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∫ab[f(x)+g(x)] dx=\int_a^b [f(x) + g(x)]\,dx =
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∫abf(x) dx+∫bcf(x) dx=\int_a^b f(x)\,dx + \int_b^c f(x)\,dx =
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∫01x2 dx=\int_0^1 x^2\,dx =
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Frequently Asked Questions

What is the Riemann integral?

The Riemann integral is defined as the limit of Riemann sums as the partition becomes finer. A function f is Riemann integrable on [a,b] if the limit exists and is independent of the choice of sample points.

Why is FTC so important?

FTC connects the two main operations of calculus: differentiation and integration are inverse operations. This lets us evaluate definite integrals using antiderivatives instead of limits of Riemann sums.

What is the Newton-Leibniz formula?

If F is an antiderivative of f on [a,b], then ∫ₐᵇ f(x)dx = F(b) - F(a). This is the fundamental connection between definite and indefinite integrals.

What are the properties of definite integrals?

Key properties include: linearity (∫[f+g] = ∫f + ∫g), additivity (∫ₐᵇ + ∫ᵇᶜ = ∫ₐᶜ), reversal (∫ₐᵇ = -∫ᵇᵃ), and monotonicity (if f ≤ g, then ∫f ≤ ∫g).

Can the lower limit also vary?

Yes! Use ∫ₐˣ = -∫ˣₐ. For ∫_{g(x)}^{h(x)} f(t)dt, differentiate as: f(h(x))h'(x) - f(g(x))g'(x).