Integration

Area under a Function Graph 2D 3D Contents

Integrals are interpreted geometrically as the area under a function graph.

Specifically, for some variable x, a function f(x), and constants a and b, the integral

∫_a^bf(x)dx

is interpreted as the area between the x-axis and portions of the graph of f(x) for which y > 0 minus the area between the x-axis and portions of the graph of f(x) for which y < 0.

[D]

Exercises

Use the demo to estimate the values of the following integrals:

∫₀¹1dx
∫₀¹xdx
∫_-1¹x²dx
∫₀¹-1dx
∫_-1¹x³dx

Left and Right Integrals 2D 3D Top of Page Contents

We can approximate the area underneath the graph of a function with a sequence of rectangles.

We can estimate the area under the graph of a function f(x) over a domain a ≤ x ≤ b by subdividing the interval into n subintervals of length (b-a)/n to get a = x_0 < x_1 < ... < x_n-1 < x_n = b where x_i = a + i(b-a)/n. We obtain the "Left Integral if f with n Subdivisions", denoted by Lf(n) = Σ_i=1ⁿf(x_i-1)(b-a)/n. Similarly we define the "Right Integral of f with n Subdivisions" to be Rf(n) = Σ_i=0^n-1 f(x_i)(b-a)/n.

[D]

Riemann Integrals

Since the Riemann integral uses the limits of these functions as n goes to infinity, the left and right integrals are the same in the limit for integrable functions. This limit is the Riemann integral. In fact we can use rectangles with height determined by the value of the function at any point in the interval [x_i, x_i+1], and the distance between subsequent x_i's may not necessarily be constant.

[D]

Exercises

1. Compare (using <, >, =) the values of the Riemann Integral for ∫_-1¹x²dx for:

1 subdivision
5 subdivisions
20 subdivisions

2. Compute the Riemann integral for ∫₀¹x³dx for three subdivisions, and use the lab to verify that your result is reasonable.

Area between Function Graphs 2D 3D Top of Page Contents

Consider two functions of one variable f(x) and g(x), and suppose we want to find the area of the region between the graphs of f(x) and g(x). Again, we can approximate the area by filling in the region with a sequence of rectangles.

These rectangles can be obtained by using the rectangles in a Riemann sum of f(x) and subtracting the rectangles in a Riemann sum of g(x) that uses the same partition. Taking the limit of these partitions, we can write this in integral form as A = ∫_a^b f(x) dx - ∫_a^b g(x) dx, or equivalently, A = ∫_a^b (f-g)(x) dx.

[D]

Exercises

1. Estimate the area between the function graphs for the example that the demo above begins with.

2. Estimate the area between the graphs f(x) = x² and g(x) = 0. What integral does this represent?

3. How can you convert a problem involving area between function graphs into a problem involving the area under a single function graph?

Arc Length for Function Graphs 2D Top of Page Contents

We can approximate the arc length of the function curve for a function f(x) by breaking the curve into segments of tangent lines.

A segment that starts at the point (x, f(x)) and ends a distance dx to the right will end at the point (x + dx, f(x + dx)) = (x + dx, f(x) + f'(x)dx). This means the length of the segment is √(dx²+(f'(x)dx)²). This expression simplifies to √(1 + (f'(x))²)dx.

Integrate with respect to x from a to b, resulting in the following formula for arc length s:

s = ∫_a^b√(1 + (f'(x))²)dx

[D]

Exercises

1. Use the demo to relate (using <, >, =) the arc lengths from -1 to 1 of the functions f(x) = 0, f(x) = x, f(x) = x², and f(x) = x³. Explain from the formula for arc length why the comparison turns out this way.

2. How would one go about using the formula for arc length to find the arc length of a function which is not continuously differentiable (e.g. f(x) = |x|)?

Change of Variables 2D 3D Top of Page Contents

The use of substitution in integration requires an application of the Change of Variables Theorem.

Suppose we want to evaluate some integral

∫_a^bf(u)du.

If this is a difficult integral to evaluate, we can often simplify by subsituting some function x(u) for part of the function f(u). For example, if f(u) = sin(2u), we can let x(u) = 2u, resulting in f(u) = sin(x(u)).

This gives the integral ∫_a^bf(x(u))du. There are two adjustments we need to make. Firstly, since u goes from a to b, x(u) must go from x(a) to x(b). Secondly, in order to integrate with respect to x, we need some way to replace du with dx. We can do this by multiplying du by dx/du and dividing the rest of the integrand by dx/du (this entire operation is equivalent to multiplying by 1).

The resulting formula for change of variables is

∫_a^bf(u)du = ∫_x(a)^x(b)[f(x(u))/x'(u)]dx.

(Note that most introductory calculus texts will write ∫_a^bf(x)dx = ∫_u(a)^u(b)[f(u(x))/u'(x)]du (hence the term "U subsitution"). The roles of x and u are switched here to make this section parallel to the two- and three- variable calculus sections.)

[D]

Exercises

1. Observe what happens for f(u) = sin(x(u)), x(u) = 1/u, u = 0.5 to 1. The two integrals shown appear to be opposites of each other, but are in fact equal. Explain why this is so.

2. Write a simplified change of variables theorem for the special case that x(u) = a * u for some constant a. Try a few examples in the demo and describe the geometric significance of this type of substitution.

Total Mass of an Interval 2D Top of Page Contents

The mass of a one-dimensional object is calculated by integrating its density function over the interval of the domain that it occupies.

For a density function ρ(x) of one variables and an object in the domain of ρ occupying a region R, the mass is

∫_Rρ(x)dx.

[D]

Exercises

Find the mass of the interval 0 ≤ x ≤ 1 for the density function ρ(x) = x³.

What is a reasonable constraint on the range of the function ρ(x)?

Center of Mass 2D 3D Top of Page Contents

The center of mass of an interval is a weighted average of the positions of the particles that the interval comprises. The amount each particle counts in this weighted average is proportional to its mass.

If we wanted to find the average position of a series of particles on an interval, we could find this position by adding up the values of the x-coordinate from each point (the only coordinate in one dimension) and dividing by the number of points. This is expressed as follows:

(Σ_nx_n)/n

Where x_n is the x coordinate of the position of the nth particle.

Suppose instead we wanted to find the center of mass of an interval as described in the first paragraph. Our expression for the x coordinate of the center of mass, given the mass m_n of each particle is:

(Σ_nx_nm_n)/(Σ_nm_n)

(Note that the denominator is equal to the total mass of the set of particles.)

Given a linear density function ρ(x), we can express this summation as

(Σ_nx_n*ρ(x_n)*h_n)/(Σ_nρ(x_n)*h_n)

Where h_n is the width of the nth particle.

For an interval with domain D composed of a very large number of particles, we can change this summation to a double integral, replacing h_n with dx:

X = (∫_Dx*ρ(x)dx)/(∫_Dρ(x)dx).

This gives us the center of mass (X) of D.

[D]

Exercises

Find a general formula for the center of mass of a wire that goes from x = 0 to x = 1 with density function ρ = xⁿ , n ≥ 0.