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Calculus: Integrals (Part 4 of x)

Preface
This series is aimed at providing tools for an electrical engineer to analyze data and solve problems in design. The focus is on applying calculus to equations or physical systems.

Introduction
This article will cover integrals.

There are many calculus references, the one I like to use is Calculus by Larson, Edwards and Hostetler.

This also assumes you are familiar with Python or can stumble your way through it.

Concepts
None

Definition of an Integral
The integral of f on the closed interval [a,b] is given by
limΔ->0 Σni=1 f(ci) Δxi = ∫ab f(x)dx = F(b)-F(a) = A, the area under the curve on the interval
provided the limit exists.

Therefore integrability implies continuity. But continuity does not imply integrability.

Properties of Integrals

General Form ∫ u dx
Constant ∫ c dx = cx + C
Sum/Difference ∫ [u +/- v] dx = ∫ u dx +/- ∫v dx + C
Chain ∫ f(u) du = F(u) + C
General Power ∫ un du = un+1 / (n+1) + C
By Parts ∫(u dv) = uv - ∫(v du) + C

Where u and v are a functions, c and n are constants.

Integral Table
Here.

Integrals of Vector Functions
If r is vector-valued function r(t) = f(t)i + g(t)j + h(t)k, then

r(t) = [∫ f(t) dt] i + [∫ g(t) dt] j + [∫ h(t) dt] k

Average Value of a Function
The average value of a function is given by the following formula:

[1/(b-a)] ∫ab f(x)dx

Initial Conditions

  1. Integrate the function
  2. Substitute initial condition and solve for C.
  3. If integrating again return to first step and repeat until finish.

An example of this is integrating the acceleration constant to determine velocity. You can substitute known initial position values to calculate the constants. Integrate again to get the position function. Then you have specific solutions for your problem.

Multiple Integration
The rules for integrating more than once are identical to a single integral, you just have to observe two things:

  1. The order of operation is inside out. The innermost integral corresponds to the innermost integration variable (e.g. dx, dy or dz). You must have the same number of integrals and integration variables.
  2. Each integration is partial. That is you hold all other variables as constants when integrating.

Volume: Disc Method
Find the volume of a function rotated about an axis and perpendicular to the axis of revolution.

Horizontal Axis of Revolution
V = π ∫ba f(x)2dx

Vertical Axis of Revolution
V = π ∫dc f(y)2dy

Volume: Shell Method
Find the volume of a function rotated about an axis and parallel to the axis of revolution.

Horizontal Axis of Revolution
V = 2π∫ba p(y)h(y)dy

Vertical Axis of Revolution
V = 2π∫dc p(x)h(x)dx

Volume: Multiple Integrals
If f is defined on a closed, bounded region R, then the double integral of f over R is:

R ∫ f(x,y) dA, where dA=dx dy.

If f(x,y,z) is continuous over a bounded solid region Q, then the triple integral of f over Q is:

∫ ∫Q ∫ f(x,y,z) dV, where dV=dx dy dz.

Arc Length
Let the function given by y=f(x) be a continuous function on interval [a,b]. The arc length s of f between a and b is:

s = ∫dc sqrt[1+f'(x)2]dx

s = ∫ba sqrt[x'(t)2+y'(t)2+z'(t)2] dt = ∫ba abs(r'(t)) dt

Surface Area: Surface of Revolution
If a smooth curve C given by x=f(t) and y=g(t), then the area S of the surface is given by

S = 2π∫bag(t) sqrt((dx/dt)2+(dy/dt)2)dt (revolved by x axis)
S = 2π∫baf(t) sqrt((dx/dt)2+(dy/dt)2)dt (revolved by y axis)

Surface Area: Multiple Integrals
If f(x,y) and its first partial derivatives are continuous on the closed region R, then the area of the surface S over R is:

S = ∫R ∫ sqrt[1+ fx(x,y)2+fy(x,y)2] dA

Relationship between derivatives and integrals

  1. Derivatives look at the tangent line and is concerned with the relationship of dy/dx (division); while the Integral looks at the area and is concerned with the relationship of dy*dx (multiplication).
  2. The integral is the antiderivative of a function. One operation is the inverse of the other.
  3. (d/dx) ∫xa [f(t)dt] = f(x)

Next Up
Vector math

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