**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*lim _{Δ->0} Σ^{n}_{i=1} f(c_{i}) Δx_{i} = ∫_{a}^{b} f(x)dx = F(b)-F(a).*

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 | ∫ u^{n} du = u^{n+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)] ∫ _{a}^{b} f(x)dx*

**Initial Conditions**

- Integrate the function
- Substitute initial condition and solve for C.
- 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:

- 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.
- Each integration is
*partial*. That is you hold all other variables as*constants*when integrating.

**Surface Area: In the Plane**

For a in a plane (flat surface) and bound on four sides by equations f(x), g(x), m(x) and n(x), the area is given by:

A = ∫_{f(x)}^{g(x)} ∫_{n(x)}^{m(x)} dy dx

**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 = ∫_{S} ∫ f(x,y,z) dS = ∫_{R} ∫ f(x,y,g(x,y)) dS

where

dS = sqrt[1+ f_{x}(x,y)^{2}+f_{y}(x,y)^{2}] dA

If **r(u,v)** = x(u,v)**i** + y(u,v)**j** + z(u,v)**k** and *D* be a region in (u,v), then the surface area surface S:

S = ∫_{S}∫ f(x,y,z) dS = ∫_{D}∫ f( x(u,v), y(u,v), z(u,v) ) dS

where

dS = abs(**r _{u}** cross

where**r _{u}** = dx/du (partial derivative)

If **r(u,v)** = x**i** + y**j** + f(x,y)**k** you will derive the same equation as above.

If **r(u,v)** = u**i** + f(u)*cos v**j** + f(u)*sin v**k** you will derive the same equation as surface of revolution.

S = ∫_{S}∫ **F ⋅ N** dS = **F ⋅ (r _{u} cross r_{v})** dA = ∫R∫

**Surface Area: Surface of Revolution**

If a smooth curve C given by y=f(x), then the area S of the surface is given by

S = 2π∫^{b}_{a}f(x) sqrt(1+(f'(x))^{2}) dx (revolved by x axis)

And for x=g(y)

S = 2π∫^{b}_{a}g(y) sqrt(1+(g'(y))^{2}) dy (revolved by y axis)

**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 ∫^{b}_{a} f(x)^{2}dx

*Vertical Axis of Revolution*

V = π ∫^{d}_{c} f(y)^{2}dy

**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∫^{b}_{a} p(y)h(y)dy

*Vertical Axis of Revolution*

V = 2π∫^{d}_{c} 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.

**Length: 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 = ∫^{d}_{c} sqrt[1+f'(x)^{2}]dx

For a vector valued function **r(t)**

s = ∫^{b}_{a} sqrt[x'(t)^{2}+y'(t)^{2}+z'(t)^{2}] dt = ∫^{b}_{a} abs(**r'(t)**) dt

For a linear (straight line through origin) this function reduces to the standard m=(y2-y1)/(x2-x1) function.

**Length: Line Integral**

This is for weighted functions. Evaluating the line integral for an unweighted function (f(x,y,z)=1) gives you the arc length above.

Let f be continuous in a region containing a smooth curve C. If C is given by **r(t)** = x(t)**i** + y(t)**j** + z(t)**k** then

s = ∫_{C} f(x,y,z) ds = ∫^{b}_{a} f(x(t), y(t), z(t)) ds

where ds = abs(**r'(t)**) dt = sqrt[x'(t)^{2}+y'(t)^{2}+z'(t)^{2}] dt

∫_{C}**F ⋅ dr** = ∫_{C}**F ⋅ T** ds = ∫^{b}_{a}**F**(x(t), y(t), z(t)) ⋅ **r'(t)** dt

∫_{C} 1 ds = ∫^{b}_{a} abs(**r'(t)**) dt

**Special Relationships***Fundamental Theorem of Line Integrals*

If **F(x,y)**=M**i**+N**j** then

∫_{C}**F ⋅ dr** = ∫_{C} ∇f ⋅ **dr** = f(x(b),y(b)) - f(x(a),y(a))

where f is a potential function of **F**. In other words for conservative functions the line integral is simply the difference between the potential function of the endpoints and is INDEPENDENT OF PATH.

*Divergence Theorem*

Let Q be a solid region bounded by a closed surface S oriented by a unit normal vector directed outward from Q. If F is a vector field whose component functions have continuous partial derivatives in Q then

∫_{S}∫ **F ⋅ N** dS = ∫∫_{Q}∫ div **F** dV

Both integrals describe total flux. The one on the left evaluates flux across the surface and the one on the right describes flux through the volume. **N** is the unit normal vector.

*Stokes's Theorem*

Let S be an oriented surface with unit normal vector **N**, bounded by a piecewise smooth simple closed curve C. If **F** is a vector field whose component functions have continuous partial derivatives on an open region containing S and C then

∫_{C}**F ⋅ dr** = ∫_{S}∫ **(curl F) ⋅ N** dS

This theorem calculates the circulation of the equation about a point.

∫_{C}**F ⋅ dr** f(x)dx = ∫_{C} ∇f ⋅ **dr** = f(x(b),y(b))-f(x(a),y(a)) where **F(x,y)** = ∇ f(x,y)

**Relationship between derivatives and integrals**

- 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). (As a matter of fact, for a straight line the derivative and integral reduce to division and multiplication, respectively.)
- ∫
_{a}^{b}f(x)dx = F(b)-F(a) - (d/dx) ∫
^{x}_{a}[f(t)dt] = f(x) - The integral is the antiderivative of a function. One operation is the inverse of the other.
- If your function is conservative then the integral of the line is
*path independent*. The line integral of a closed curve is always*zero*.

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