# 9.1 Computing the Area Under a Curve

- Page ID
- 9464

This chapter essentially deals with the problem of computing the area under a curve. First, we will employ a basic approach and form trapezoids under a curve. From these trapezoids, we can calculate the total area under a given curve. This method can be tedious and is prone to errors, so in the second half of the chapter, we will utilize a built-in MATLAB function to carry out numerical integration.

### A Basic Approach

There are various methods to calculating the area under a curve, for example, __Rectangle Method__, __Trapezoidal Rule__ and __Simpson's Rule__. The following procedure is a simplified method.

Consider the curve below:

Figure \(\PageIndex{1}\). Numerical integration

Each segment under the curve can be calculated as follows:

\(\frac{1}{2}\left(y_{0}+y_{1}\right) \Delta x+\frac{1}{2}\left(y_{1}+y_{2}\right) \Delta x+\frac{1}{2}\left(y_{2}+y_{3}\right) \Delta x\)

Therefore, if we take the sum of the area of each trapezoid, given the limits, we calculate the total area under a curve. Consider the following example.

Given the following data, plot an x-y graph and determine the area under a curve between x=3 and x=30

Index |
x [m] |
y [N] |

1 | 3 | 27.00 |

2 | 10 | 14.50 |

3 | 15 | 9.40 |

4 | 20 | 6.70 |

5 | 25 | 5.30 |

6 | 30 | 4.50 |

Data Set

First, let us enter the data set. For x, issue the following command `x=[3,10,15,20,25,30];`

. And for y, `y=[27,14.5,9.4,6.7,5.3,4.5];`

. If yu type in `[x',y']`

, you will see the following tabulated result. Here we transpose row vectors with ' and displaying them as columns:

ans = 3.0000 27.0000 10.0000 14.5000 15.0000 9.4000 20.0000 6.7000 25.0000 5.3000 30.0000 4.5000

Compare the data set above with the given information in __the question__.

To plot the data type the following:

```
Callstack:
at (Under_Construction/Book:_A_Brief_Introduction_to_Engineering_Computation_with_MATLAB_(Beyenir)/09:_Numerical_Integration), /content/body/section/div[1]/div/div/div[1]/pre, line 3, column 18
```

The following figure is generated:

Figure \(\PageIndex{2}\). Distance-Force Graph

To compute dx for consecutive x values, we will use the index for each x value, see the given data in __the question__.:

dx=[x(2)-x(1),x(3)-x(2),x(4)-x(3),x(5)-x(4),x(6)-x(5)];

dy is computed by the following command:

dy=[0.5*(y(2)+y(1)),0.5*(y(3)+y(2)),0.5*(y(4)+y(3)),0.5*(y(5)+y(4)),0.5*(y(6)+y(5))];

dx and dy can be displayed with the following command: `[dx',dy']`

. The result will look like this:

```
Callstack:
at (Under_Construction/Book:_A_Brief_Introduction_to_Engineering_Computation_with_MATLAB_(Beyenir)/09:_Numerical_Integration), /content/body/section/div[1]/div/div/div[2]/div/div[2]/pre, line 3, column 5
```

Our results so far are shown below

x [m] |
y [N] |
dx [m] |
dy [N] |

3 | 27.00 | ||

10 | 14.50 | 7.00 | 20.75 |

15 | 9.40 | 5.00 | 11.95 |

20 | 6.70 | 5.00 | 8.05 |

25 | 5.30 | 5.00 | 6.00 |

30 | 4.50 | 5.00 | 4.90 |

x, y and corresponding differential elements

If we multiply dx by dy, we find da for each element under the curve. The differential area da=dx*dy, can be computed using the 'term by term multiplication' technique in MATLAB as follows:

da=dx.*dy da = 145.2500 59.7500 40.2500 30.0000 24.5000

Each value above represents an element under the curve or the area of trapezoid. By taking the sum of array elements, we find the total area under the curve.

sum(da) ans = 299.7500

__The following__ illustrates all the steps and results of our MATLAB computation.

x [m] |
y [N] |
dx [m] |
dy [N] |
dA [Nm] |

3 | 27.00 | |||

10 | 14.50 | 7.00 | 20.75 | 145.25 |

15 | 9.40 | 5.00 | 11.95 | 59.75 |

20 | 6.70 | 5.00 | 8.05 | 40.25 |

25 | 5.30 | 5.00 | 6.00 | 30.00 |

30 | 4.50 | 5.00 | 4.90 | 24.50 |

299.75 |

Computation of the approximate area under a curve

### The Trapezoidal Rule

Sometimes it is rather convenient to use a numerical approach to solve a definite integral. The trapezoid rule allows us to approximate a definite integral using trapezoids.

The

trapz

#### Command

Z = trapz(Y) computes an approximation of the integral of Y using the trapezoidal method.

Now, let us see a typical problem.

Given Area \(=\int_{2}^{5} x^{2} d x\) , an analytical solution would produce 39. Use trapz command and solve it

- Initialize variable x as a row vector, from 2 with increments of 0.1 to 5:
`x=2:.1:5;`

- Declare variable y as
`y=x^2;`

. Note the following error prompt:`??? Error using ==> mpower Inputs must be a scalar and a square matrix.`

This is because x is a vector quantity and MATLAB is expecting a scalar input for y. Because of that, we need to compute y as a vector and to do that we will use the dot operator as follows:`y=x.^2;`

. This tells MATLAB to create vector y by taking each x value and raising its power to 2. - Now we can issue the following command to calculate the first area, the output will be as follows:
area1=trapz(x,y) area1 = 39.0050

Notice that this numerical value is slightly off. So let us increase the number of increments and calculate the area again:

x=2:.01:5; y=x.^2; area2=trapz(x,y) area2 = 39.0001

Yet another increase in the number of increments:

x=2:.001:5; y=x.^2; area3=trapz(x,y) area3 = 39.0000

Determine the value of the following integral:

\(\int_{0}^{\pi} \sin (x) \mathrm{d} x\)

- Initialize variable x as a row vector, from 0 with increments of pi/100 to pi:
`x=0:pi/100:pi;`

- Declare variable y as
`y=sin(x);`

- Issue the following command to calculate the first area, the output will be as follows:
area1=trapz(x,y) area1 = 1.9998

let us increase the increments as above:

x=0:pi/1000:pi; y=sin(x); area2=trapz(x,y) area2 = 2.0000

A gas expands according to the law, PV^{1.4}=c. Initially, the pressure is 100 kPa when the volume is 1 m^{3}. Write a script to compute the work done by the gas in expanding to three times its original volume^{1}.

Recall that PV diagrams can be used to estimate the net work performed by a thermodynamic cycle, __see Wikipedia__ or we can use definite integral to compute the work done (WD) as follows:

\(\mathrm{WD}=\int p \mathrm{d} v\)

If we rearrange the expression pressure as a function of volume, we get:

\(P=\frac{c}{V^{1.4}}\)

By considering the initial state, we can determine the value of c:

\(\begin{aligned} c &=100 \times 1^{1.4} \\ &=100 \end{aligned}\)

From __the equation__ and __the equation__ above, we can write:

\(P=\frac{100}{V^{1.4}}\)

\(\mathrm{WD}=\int_{1}^{3} \frac{100}{v^{1.4}} \mathrm{d} v\)

For MATLAB solution, we will consider __P as a function of V__ and __WD__. Now, let us apply the three-step approach we have used earlier:

- Initialize variable volume as a row vector, from 1 with increments of 0.001 to 3:
`v=1:0.001:3;`

- Declare variable pressure as
`p=100./v.^1.4;`

- Use the
`trapz`

function to calculate the work done, the output will be as follows:WorkDone=trapz(v,p) WorkDone = 88.9015

These steps can be combined in an m-file as follows:

(click for details)`Callstack: at (Under_Construction/Book:_A_Brief_Introduction_to_Engineering_Computation_with_MATLAB_(Beyenir)/09:_Numerical_Integration), /content/body/section/div[2]/div[4]/ol/li[3]/div/div/pre, line 4, column 7`

A body moves from rest under the action of a direct force given by \(F=\frac{15}{x+3}\)

where x is the distance in meters from the starting point. Write a script to compute the total work done in moving a distance 10 m.^{2}

Recall that the general definition of mechanical work is given by the following integral, __see Wikipedia__:

\(\mathrm{WD}=\int F \mathrm{d} x\)

Therefore we can write:

\(\mathrm{WD}=\int_{0}^{10} \frac{15}{x+3} \mathrm{d} x\)

Applying the steps we followed in the previous examples, we write:

```
Callstack:
at (Under_Construction/Book:_A_Brief_Introduction_to_Engineering_Computation_with_MATLAB_(Beyenir)/09:_Numerical_Integration), /content/body/section/div[2]/div[6]/div/pre, line 4, column 7
```

The output of the above code is:

A body moves from rest under the action of a direct force given by F=15/(x+3) where x is the distance in meters From the starting point. Compute the total work done in moving a distance 10 m. WorkDone = 21.9951

The

integral

### Function

As we have seen earlier, `trapz`

gives a good approximation for definite integrals. The `integral`

function streamlines numerical integration even further. Before we learn about `integral`

function, first we will look at anonymous functions.

#### Anonymous Functions

An anonymous function is a function that can be defined in the command window (i.e. it does not need to be stored in a program file). Anonymous functions can accept inputs and return outputs, just as standard functions do such as `sqrt(X)`

or `log(X)`

.

To define an anonymous function, first we create a handle with `@(x)`

and type in the function: `myfunction=@(x) x^2+1`

.

If you want to evaluate `myfunction`

at 1, just type in `a=myfunction(1)`

at the command window and you get the result of `2`

.

**Syntax for **** integral**To evaluate an integral from a minimum to a maximum value, we specify a function and its minimum and maximum

`Z = integral(fun,xmin,xmax)`

.

Given `y=x^2`

, evaluate the integral from x=2 to x=5 as we have done it with __ trapz__ command.

- Define function
`myfunction=@(x) x.^2;`

- Apply the syntax to
`myfunction`

as follows`Z = integral(myfunction,2,5)`

- You should get a result of
`Z = 39`

. aside

Notice that, unlike in

trapz

example, we did not need to define a vector and change the increments to get an accurate result.

### Summary of Key Points

- In its simplest form, numerical integration involves calculating the areas of segments that make up the area under a curve,
- MATLAB has built-in functions to perform numerical integration,
`Z = trapz(Y)`

computes an approximation of the integral of Y using the trapezoidal method.- Anonymous functions are inline statements that we can define with
`@(x)`

, `Z = integral(fun,xmin,xmax)`

numerically integrates function`fun`

from`xmin`

to`xmax`

.