Integration

Engineering Context

Integration is widely used across engineering disciplines, both as a direct means to solve a problem and as a tool to solve more complex problems. It can be used to determine areas and volumes of irregular shapes, to find position based on acceleration or velocity, to find work done by an inconsistent force, etc.

MAE:
Integration can be used in mechanical engineering to determine the work done by a force across a certain distance. For example, say a mechanical engineer wants to determine the work that is done by a car’s engine when driving on a level surface. Work is the integral of force, so they could calculate work using the equation

\[ W = \int F(x) \, dx \]

Where \( W \) is work and \( F(x) \) is force produced by the engine at a given position \( x \). Say the force is inconsistent, and its value at point \( x \) can be found with the function \( F(x) = 2x \). Then \( W = \int 2x \, dx = x^2 + C \). By inputting some position value \( x \) into this function \( W \), the work done up to that point can be found.

ECE:
Integration is used in electrical engineering to find the voltage across a capacitor when a current is passed through it. It can also be used to find total stored energy of a capacitor. For example, the power supplied to a capacitor is given by the equation:

\[ p(t) = \frac{dw(t)}{dt} \]

Integrating this equation can help you find \( w(t) \), the stored energy in a capacitor.

BENG:
Biological engineers often work with enzymatic or chemical reactions, and want to know the rates at which these reactions proceed. Integration is heavily used in this study of chemical kinetics. For example, the rate of a simple chemical equation is given by

\[ \frac{d[A]}{dt} = \text{rate} = k[A]^n \]

where [A] is the concentration of the reactant in \(\frac{mol}{L}\); \( n \) is the order of the reaction and can have values of 0, 1, 2, and occasionally 3; and \( k \) is some rate constant that varies depending on the reaction. Essentially what this equation is telling us is that, in a given chemical equation, the rate at which a reactant reacts over time is proportional to how much of that reactant is already present. Typically, when the initial concentration of the reactant is very large, the reaction will proceed at a relatively faster rate. This is all well and good, but say a biological engineer wants to know not just how fast a reaction proceeds, but how much of a certain reactant will be present at a given time. They can determine this by integrating the above rate law (in this case assuming that the reaction order is 1):

\[ \frac{d[A]}{dt} = k[A]^1 \]
\[ \frac{d[A]}{dt} * \frac{1}{[A]} = k \]
\[ \int \frac{d[A]}{dt} * \frac{1}{[A]} dt = \int k dt \]
\[ \ln[A] = kt + C, \quad [A] = Ce^{kt}, \quad C = A_0 \]

The rate equation is now in terms of time, and the concentration of reactant A can be determined by inputting a certain time \( t \).

CEE:
As a civil or environmental engineer, it is often necessary to calculate areas and volumes of materials required for a project. Integration is a good method for calculating areas simply and quickly. For example, say a civil engineer is working on designing a road that goes over a steep hill. They need to cut some material from the top of the hill so that the road is not dangerously steep. Say the scaled profile of the hill approximately fits the graph of −1 2x2 + 2 (red line), and the desired profile approximately fits the graph of −1 4x2 + 1 (blue line):

CEE example graph

The engineer can calculate the volume that needs to be cut away using the following steps:

  1. Compute \( \int_{-2}^{2} (x^2 + 2) \, dx \) to find the area of the hill’s profile
  2. Compute \( \int_{-2}^{2} \left( \frac{1}{4}x^2 + 1 \right) \, dx \) to find the desired area
  3. Subtract the second integral from the first to find the difference between the two areas
  4. Multiply the difference by the width of the roadway to calculate the total volume that needs to be cut

The Essentials

An integral is a function that finds the area under a given curve. It is the inverse operation to a derivative. An integral can be definite, meaning it has distinct starting and ending points, or indefinite, meaning it does not. Integrals appear in the form

\[ \int_{a}^{b} f(x) \, dx \]

where a and b are the lower and upper bounds of integration, f(x) is the function being integrated, and dx tells us that the integral is being taken with respect to the variable x.

To understand an integral conceptually, it can help to think of it as a sum of many small area sections under a curve:

The concept of an integral -the sum of many small area sections under a curve

In the figure above, you can see that the sum of all of these rectangles will be approximately equal to the area under the curve, and that as the width of the rectangles gets smaller and smaller, the sum will be an increasingly accurate approximation. An integral performs essentially this exact same function of summing rectangles, but it takes an infinite sum of infinitely small partitions of x, or infinitely thin rectangles, in order to find the exact area under the curve.

Mathematically, taking the integral of a function \( f(x) \) returns another function, \( g(x) \). If you plug some x into \( g(x) \), it will give you the value of the area under f(x) from 0 to x:

graph explaining an integral

When you take an indefinite integral of a function \( f(x) \), the output of the integral will be another function, namely the antiderivative of \( f(x) \), plus some unknown constant \( C \):

\[ \int 3x^2 \, dx = x^3 + C \]
\[ \int \frac{1}{x} \, dx = \ln(x) + C \]
\[ \int \cos(x) \, dx = \sin(x) + C \]

These examples help us see that an integral is the inverse function to a derivative. For example, the derivative of \( x^3 \) plus some constant \( C \) is \( 3x^2 \), and thus the integral of \( 3x^2 \) is \( x^3 + C \). This \( + C \) is important because it represents a constant that may have been in the original equation before we took the derivative. Taking the derivative would have caused it to disappear. For example, \( 3x^2 \) is the derivative of a whole family of functions: \( x^3 + 5 \), \( x^3 + 2 \), \( x^3 + \pi \), or any function of the form \( x^3 + C \).

A definite integral is essentially a bounded form of an integral. It will appear in the same form as an indefinite integral, but it has set values for a and b, the upper and lower bounds of integration. Here is an example:

\[ \int_{1}^{3} x^2 \, dx \]

Taking a definite integral of a function \( f(x) \) involves taking the antiderivative \( g(x) \) of the function and then evaluating it using given parameters. The value of the definite integral is \( g(b) - g(a) \). Using the above example where \( f(x) = 3x^2 \) and \( g(x) = x^3 \):

\[ \int_{1}^{3} 3x^2 \, dx = x^3 \bigg|_{1}^{3} = (3)^3 - (1)^3 = 8 \]

Notice that you only add a + C when you are evaluating indefinite integrals. In summary, evaluating an indefinite integral yields another function, while evaluating a definite integral typically yields some distinct numerical value.

To compute basic integrals, we can use our knowledge of derivatives and just work backwards. For example, since we know that the derivative of \( \sin(x) \) is \( \cos(x) \), we can know that the integral of \( \cos(x) \) is \( \sin(x) \). When taking the integral of polynomials, we can use a sort of inverse of the power rule for derivatives: \( \int ax^n \, dx = \frac{a}{n+1} x^{n+1} + C \)

When we get into more complex integrals, such as integrals of products or quotients, different methods and techniques of integration come in handy. (See the sections on integration rules, substitution, and integrals involving exponential and logarithmic functions to learn more about these techniques.)

A Deeper Dive

Taking a Riemann sum of a curve can help us to visualize an integral. A Riemann sum approximates the area under a curve using the formula:

\[ A \approx \sum_{i=1}^{n} f(x_i^*) \Delta x \]

on a closed interval \([x_a, x_b]\). \( n \) refers to the number of subintervals the interval is divided into, \( i \) is the designation of each subinterval, \( x_i^* \) a point anywhere on each subinterval, and \( \Delta x \) is the width of each subinterval. Essentially, what this function is doing is summing up a series of rectangles under the curve, from the point \( x_a \) to the point \( x_b \):

visual representation of a right-hand Riemann sum

Figure: visual representation of a right-hand Riemann sum of \( y = \frac{1}{2}x^2 - 2x + 2 \) on interval \([0, 2]\) with \( n = 10 \)

To get from a Riemann sum to an integral, take the limit of the sum as n goes to infinity:

\[ A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_{a}^{b} f(x) \, dx \]

As \( n \), the number of subintervals, goes to infinity, \( \Delta x \) will get smaller and smaller in value until it is the infinitely small differential \( dx \). At this point, the sum is exactly equal to the area under the curve, and thus is the integral of the curve. It is important that this differential value \( dx \) stay in the notation of an integral so we know which variable we are integrating with respect to – in this case, that we are taking the infinite sum of a function by summing infinitely small partitions of \( x \). This notation comes in particularly handy when performing variable substitutions to compute integrals.

Practice

Problems

  1. Compute the following indefinite integrals:

    1. \( \int x^3 + 5x \, dx \)
    2. \( \int \cos(x) \, dx \)
    3. \( \int x^{\frac{1}{2}} \, dx \)

  2. Compute the following definite integrals:

    1. \( \int_{0}^{4} 3x^2 + 4x \, dx \)
    2. \( \int_{0}^{\pi} \sin(x) \, dx \)
    3. \( \int_{-2}^{2} \frac{x}{2} \, dx \)

Solutions:

  1. Indefinite integrals:

    1. \( \frac{1}{4}x^4 + \frac{5}{2}x^2 + C \)
    2. \( \sin(x) + C \)
    3. \( \frac{2}{3}x^\frac{3}{2} + C \)

  2. Definite integrals:

    1. 96
    2. 2
    3. 0

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