Integral+Applications

In calculus, one may encounter the use of Integrals or Anti-Derivatives. But how are these applied to real-world scenarios?
 * //__ The Application of Integrals __//**

There are many ways to find areas of shapes. For example, it is well known that you can find the volume of a box by multiplying its length by its width and its height. However, sometimes the volumes we need to compute are not so simple. For example, how could we calculate the exact volume inside of a recyclable water bottle? Or that of an old vase? This is where integrals come into play. Let's start with the basics:

//__ Part 1:Areas through integration __//

// I: Finding the Area Under a Curve // On a basic level, integrals can be used just to find the area of shapes. Say you wanted to find the area, S, under the following curve from a to b:

This area can be found with a simple standard integral:, where S is the area, a and b are limits of integration that show the interval on which the area is to be found, and f(x) is the equation of the curve you are integrating. This standard format can be used to find the area of any single continuous curve.

The idea of finding an area under a curve helps people to find areas for odd shapes, such as layouts for certain cuts of paper, tile, rocks, clothing, etc.

// II: Finding the Area Under Jagged Shapes or a System of Curves // Sometimes, curves are bent or connected to another line or curve:
 * [[image:Added Areas.jpg]]

Sometimes, odd-shaped curves have one set equation that accounts for all the bends and twists it contains. However, some can be dismantled and taken in chunks to make integration simpler. As you can see, the function y=f(x) above can be broken into three parts: a section with a negative derivative on the interval [0,1], one with a positive derivative on the interval [2,3], and one with a derivative of 0 on the interval [3,5].

An easy way to find the area of the curve as a whole would be to find the areas under each of the slopes respectively. The first portion of the graph can be called, the second, and the third. Judging by the graph, it can be determined that the equations of each of the line segments are, , and respectively.

In order to find the area under the entire graph, one can add up individual areas under the graph together. In this case, you could use the following integral to do just that: The limits of integration come from where each slope changes. The addition of the partial areas gives you the total area under the changing graph.

// III: Finding the Area Between Curves // The application of integrals stems from the idea of sums of different distances with set limits. The “area under the graph” idea is applied to a new set of areas, including differences between sets of integrals and three-dimensional volume calculations.

Sometimes the areas people look to find do not have a set type of formula to follow, like the area of a square, triangle, or circle do. When trying to find the area of “odd” shapes or groups of curves and lines, one could do so by using multiple integrals. For example, imagine having curves f(x) and g(x), where as shown below:

Because the functions are in terms of x, our anti-derivatives (integrals) need to be in terms of dx. In a situation like this, we are solving for the area by finding the sum of all the vertical distances between points on the two curves. Simply put, the idea of integral subtraction is to subtract a small area from a large one to get the difference between them. This can be expressed through the subtraction of the lower integral from the higher one, or:

where g(x) is less than f(x)

If the function set was similar to the following:

then the integral could be solved in terms of y. Because the functions are in terms of y, our anti-derivatives (integrals) need to be in terms of dy. In a situation like this, we are solving for the area by finding the sum of all the horizontal distances between points on the two curves. Again, the idea of integral subtraction is to subtract a small area from a large one to get the difference between them. This can be expressed through the subtraction of the innermost integral from the outermost one, or: where g(y) is closer to the vertical axis than f(y) is.

But how do you get an equation in terms of y? Say you are given a question asking you to find the area between the curves and in terms of dy. In order to construct a correct integral in that format, it is best to change each equation to be in terms of y (or set equal to x) as shown below: Notice how each of the equations is in terms of y. The limits of integration will come from the maximum and minimum y values. In this case, the interval would be from -2 to 4 on the y axis. With this being said, the integral expression would be something like



//__ Part 2: From 2-D to 3-D __//

// I: Cross Sectional Areas // In some cases, certain shapes must be extruded from previously drawn curves or lines in order to make a three-dimensional object. For example, if you were told to imagine the following curves with square cross sectional areas, the visual you could get could roughly be resembled by: To find the volume of the shape formed by this extrusion, one would need to do an integral from a to b using as many square slices possible. In order to do this, one must know how to find the area and volume of each slice. On a basic level, the area of a square is represented by, meaning the area is equal to the length of any side squared. The volume of a single square slice of the above graph could be seen as, meaning the area of the squared multiplied by an applied, 3-D height helps to get the volume of a single slice. To find the total volume using as many squares as possible, the distance between them needs to be minimal, represented by dx. In order to construct an approprite integral, one has to apply the Volume of a slice idea: As you can see, the volume of a slice idea remains present. The f(x) represents the length of a single side from the axis to the curve itself. The dx represents every single possible distance between each slice. Once integrated, you will have found the volume of something with square cross sections.

The same idea can be applied to volumes with cross sections of different shapes: half-circles, rectangles, triangles, and basically any polygon. The important thing to remember is the concept: you apply the volume per slice (a.k.a. the area of each shape type multiplied by a height) and then translate it into Integral speak. For a half-circle, the radius would be half the distance between the function and axis. This same idea may be applied to the areas between two curves or integrals along the y-axis as well.

// II: The Disk Method // Sometimes, people need to find the volume of a container or object that has curved or oddly-shaped sides: a barrel, vase, water bottle, propane tank, pool filter, etc. Because we already know how to find the //Area// under a curve, let us explore how to make that into a //Volume//.

In order to make three dimensional objects out of two dimensional graphs, one may use and rotate integrals around an axis to make a uniform volume. For example, if you revolve the following graph:

around the x-axis, the resulting volume will be

you can see that any slice made by cutting the shape vertically would be a perfect circle. Using this idea, the area of a slice of any fully-rotated graph would be. In order to find the area of the entire sold shape, one could apply an integral in the form of the area equation so as to say “the area of every possible slice of this shape put together”. When writing the integral, the “radius” of each disk is from the axis of revolution to the curve itself, or put simply, Radius=f(x). So, becomes

The reason the integral is written like that is to say that on the given integral from a to b, the sums of the areas of each circular slice gives you the volume of the shape from a to b. This method is known as the Disk Method solely because each cross-sectional is a perfect circle.

// III: The Washer Method // A similar idea is explored when rotating the area between two graphs around an axis. People would use this to find the volumes of things like inner tubes, the plastic used to make a funnel, or even a doughnut. For example, if you were to rotate the area between the curves and below around an axis, its cross sectional areas would look like rings at every point where the two curves do not intersect (in other words, on the interval [0,1]):
 * [[image:Area between curves to revolve.gif]]

To find the area of a washer or ring, you can find the area of the outer circle and subtract the area from the inner circle from it. Simply put:. Re-writing this to accommodate individual area equations yields Writing an integral to sum up the slices of this shape combines the ideas of finding the areas between curves and the Disk Method. In order to convert the A(washer) equation into a volume equation, an integral is needed. As is done in the Disk Method, replace individual names for “radius” with the equations given. In this case the equation would read: The integral application here is similar to that of the Disk Method in that it is used to sum up every slice on the designated integral. The idea of finding areas between curves is applied when trying to find individual areas between the circles within each washer in order to find the area of the washer. The innermost circle(in this case ) is subtracted from the outermost circle (in this case ) to get the area between them. Once the integral is applied, the individual areas of each and every possible slice on the interval [0,1] is summed and the volume is found.

//__ Part 3: Scientific Applications __//

// I: Finding the Center of Mass Under a Curve // In certain industries, namely the construction industry, large pieces of material need to be picked up and put together in order to build something. However, a majority of oddly shaped objects my crack or splinter if lifted the wrong way. This is due to differences in tolerance and durability. In order to prevent failure, objects should be lifted at their //centroids//, or center of mass, so that there is an equal distribution of weight. These have a horizontal (x) and vertical (y) component and thus have a coordinate pair.

Centroids are found based upon an object's tendency to rotate around a point, called a //moment//. Centroids of shapes with straight sizes and edges can be found using simple equations relating distance to mass. Summations of these moments in both the x and y directions placed respectively over the total area can find the coordinates in both the x and y directions of these shapes. However, let's focus on finding the centroid of shapes with curved sides.

Imagine we had to find the centroid of the following graph:

The first thing you would want to do is establish what an individual moment would be. A single moment for the above curve is a single slice, either dx or dy. Shown above is dx. One can calculate the x coordinate for the centroid of the curve above using this equation: which says that the coordinate can be found by calculating the sum of all dx's and placing them over the total area. The integral calculates the sum of the moments. The same format can be used to find the y coordinate:

// II: Finding the Center of Mass Between Two Curves // In similar cases, one can apply the idea of integral subtraction to find the coordinates of the centroid of a shape between two curves: The coordinates for the centroid can be found using the idea behind integral subtraction inside of the integral format for a centroid: and  respectively.

These centroids are the center of mass in an object, meaning that if the object was held at these points, they would be perfectly balanced. The centroids only need to be found on the area of a face of a wall solely because the centroid of the face would remain the same no matter the thickness of the wall.

// III: Work from a Variable Force // Work is defined as the measure of a force applied over a specified distance. But sometimes, the force varies as opposed to remaining constant. How can we calculate the work if one of the variables is actually changing? The answer: Integration.

When the force is the variable, it can be represented as a function of x. Let's call this function F(x) for force. The distance it is applied would be represented by dx. the equation to calculate the total work done over a certain distance interval would be.

This work equation may be applied to electric charges as well. If a single force between charges is represented by:, where f(x) represents the force between the charges, k equals a discovered constant, x is equal to the distance between two particles, and the q's are the measures of each of the charges respectively, then the work done can be calculated by adding up all of these forces on a certain interval using an integral:.

As you can see, work is found by adding up many instantaneous forces over a certain distance or time interval.

// IV:Finding Distance, Velocity or Acceleration // Another common example of an applied integral is seen when finding velocity from acceleration or distance from velocity. Say you had to find the height after 4 seconds of a ball that was thrown straight up into the air at a rate of 24.0 m/s. The height that is being asked for is the distance traveled upward, or simply, the distance. The only acceleration bringing the ball down is the acceleration of gravity (9.81 m/s2 ). To get from the data given to a distance calculation, one must apply a series of integrals. Because velocity is the derivative of distance, one can say that distance (s) is the anti-derivative of velocity (v), shown as:. To find an equation for velocity at any given point, one must anti-derive or integrate the acceleration equation, shown as:. Applying these two ideas to our specific equation, firstly you have to find an equation for acceleration. Because it was established that a=9.81m/s2, one can say that. Or. When integrating something with a variable, you add one to the power, bring the reciprocal of that number down as a coefficient, and add C as a possible constant. So, our becomes  or.

// V: Forces Due to Liquid Pressures // When objects are submerged in a liquid, the liquid exerts a force on the object. The pressure exerted by the liquid needs to be calculated for things like ships, submarines, and scuba gear so that they will not collapse/cave in when used.

The first thing to consider is how to calculate the force of something when placed under liquid. The force can be calculated as the product of an object's area (a), depth in the liquid (y), and the density of the liquid (w), shown as: This force calculation is just the force on a portion of an object.

Say we had an oddly shaped portion of a sunken ship, shown by: The force calculation helps to find the force on the tiniest of areas, A, which can be seen as dx when applied to an integral. Therefore, to sum up __all__ of the forces between the top of the piece (depth a) and the bottom (depth b) you need to do an integral, shown as:where F is the force on the entire piece, x is the varying length, w is the constant density of the liquid, y is the depth of the x below the surface, and a and b are the top and bottom of the piece. This integral is in terms of dy solely because you are layering horizontal area pieces along the vertical axis.

//__ Part 4: Integral Extras __//

// I: Finding the Average Value of a Function // Finding the average of something is relatively simple: you add all parts and divide by the number of parts. To find the average on a line you add up two points on an interval and divide by the change in distance between them. These values can be used to find a value that people use as a benchmark reference for data plots.

Sometimes, these averages must be applied over a range. For example, say you wanted to calculate the average temperature of a day using a function that recorded all the temperatures that occurred. You would want to find the area under the curve at every possible point on the function so as to add up every single possible value that passed.

To do this, one could add an interval and function to the standard equation for calculating average value: where y is the average value you are trying to find, f(x) is the curve of values, dx is the change between values, and a and b are the limits of the average.