![int(a*cos(pi*t))
int(a^x)
Whenyourunthe file, it displays the following result:
ans =
piecewise([n == -1, log(x)], [n ~= -1, x^(n + 1)/(n + 1)])
f =
sin(n*t)
ans =
-cos(n*t)/n
ans =
(a*sin(pi*t))/pi
ans =
a^x/log(a)
Example 2
Create a script file and type the following code init:
syms x n
int(cos(x))
int(exp(x))
int(log(x))
int(x^-1)
int(x^5*cos(5*x))
pretty(int(x^5*cos(5*x)))
int(x^-5)
int(sec(x)^2)
pretty(int(1 - 10*x + 9 * x^2))
int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2)
pretty(int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2))
Note that the pretty command returns anexpressionina more readable format.
Whenyourunthe file, it displays the following result:
ans =
sin(x)
ans =
exp(x)
ans =
x*(log(x) - 1)
ans =
log(x)
ans =
(24*cos(5*x))/3125 + (24*x*sin(5*x))/625 - (12*x^2*cos(5*x))/125 + (x^4*cos(5*x))/5 -
(4*x^3*sin(5*x))/25 + (x^5*sin(5*x))/5
2 4
24 cos(5 x) 24 x sin(5 x) 12 x cos(5 x) x cos(5 x)
----------- + ------------- - -------------- + ----------- -
3125 625 125 5
3 5](https://image.slidesharecdn.com/matlabintegration-150713153258-lva1-app6891/85/Matlab-integration-2-320.jpg)
![f = x;
c = [1, 0];
integral = polyint(c);
a = polyval(integral, 9) - polyval(integral, 4);
display('Area: '), disp(double(a));
Analternative solutioncanbe givenusing quad() functionprovided by Octave as follows:
pkg load symbolic
symbols
f = inline("x");
[a, ierror, nfneval] = quad(f, 4, 9);
display('Area: '), disp(double(a));
Example 1
Let us calculate the area enclosed betweenthe x-axis, and the curve y = x3−2x+5 and the ordinates x = 1 and x =
2.
The required area is givenby:
Create a script file and type the following code:
f = x^3 - 2*x +5;
a = int(f, 1, 2)
display('Area: '), disp(double(a));
Whenyourunthe file, it displays the following result:
a =
23/4
Area:
5.7500
Following is Octave equivalent of the above calculation:
pkg load symbolic
symbols
x = sym("x");
f = x^3 - 2*x +5;
c = [1, 0, -2, 5];
integral = polyint(c);
a = polyval(integral, 2) - polyval(integral, 1);
display('Area: '), disp(double(a));
Analternative solutioncanbe givenusing quad() functionprovided by Octave as follows:
pkg load symbolic
symbols](https://image.slidesharecdn.com/matlabintegration-150713153258-lva1-app6891/85/Matlab-integration-4-320.jpg)
![x = sym("x");
f = inline("x^3 - 2*x +5");
[a, ierror, nfneval] = quad(f, 1, 2);
display('Area: '), disp(double(a));
Example 2
Find the area under the curve: f(x) = x2 cos(x) for −4 ≤ x ≤ 9.
Create a script file and write the following code:
f = x^2*cos(x);
ezplot(f, [-4,9])
a = int(f, -4, 9)
disp('Area: '), disp(double(a));
Whenyourunthe file, MATLAB plots the graph:
and displays the following result:
a =
8*cos(4) + 18*cos(9) + 14*sin(4) + 79*sin(9)
Area:
0.3326
Following is Octave equivalent of the above calculation:
pkg load symbolic
symbols
x = sym("x");
f = inline("x^2*cos(x)");
ezplot(f, [-4,9])
print -deps graph.eps
[a, ierror, nfneval] = quad(f, -4, 9);](https://image.slidesharecdn.com/matlabintegration-150713153258-lva1-app6891/85/Matlab-integration-5-320.jpg)
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Matlab integration
- 1. http://www.tutorialspoint.com/matlab/matlab_integ ration.htm Copyright © tutorialspoint.com MATLAB - INTEGRATION Integrationdeals withtwo essentially different types of problems. Inthe first type, derivative of a functionis givenand we want to find the function. Therefore, we basically reverse the process of differentiation. This reverse process is knownas anti-differentiation, or finding the primitive function, or finding anindefinite integral. The second type of problems involve adding up a very large number of very smallquantities and then taking a limit as the size of the quantities approaches zero, while the number of terms tend to infinity. This process leads to the definitionof the definite integral. Definite integrals are used for finding area, volume, center of gravity, moment of inertia, work done by a force, and innumerous other applications. Finding Indefinite Integral Using MATLAB By definition, if the derivative of a functionf(x) is f'(x), thenwe say that anindefinite integralof f'(x) withrespect to x is f(x). For example, since the derivative (withrespect to x) of x2 is 2x, we cansay that anindefinite integralof 2x is x2. Insymbols: f'(x2) = 2x, therefore, ∫ 2xdx = x2. Indefinite integralis not unique, because derivative of x2 + c, for any value of a constant c, willalso be 2x. This is expressed insymbols as: ∫ 2xdx = x2 + c. Where, c is called an'arbitrary constant'. MATLAB provides anint command for calculating integralof anexpression. To derive anexpressionfor the indefinite integralof a function, we write: int(f); For example, fromour previous example: syms x int(2*x) MATLAB executes the above statement and returns the following result: ans = x^2 Example 1 Inthis example, let us find the integralof some commonly used expressions. Create a script file and type the following code init: syms x n int(sym(x^n)) f = 'sin(n*t)' int(sym(f)) syms a t
- 2. int(a*cos(pi*t)) int(a^x) Whenyourunthe file, it displays the following result: ans = piecewise([n == -1, log(x)], [n ~= -1, x^(n + 1)/(n + 1)]) f = sin(n*t) ans = -cos(n*t)/n ans = (a*sin(pi*t))/pi ans = a^x/log(a) Example 2 Create a script file and type the following code init: syms x n int(cos(x)) int(exp(x)) int(log(x)) int(x^-1) int(x^5*cos(5*x)) pretty(int(x^5*cos(5*x))) int(x^-5) int(sec(x)^2) pretty(int(1 - 10*x + 9 * x^2)) int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2) pretty(int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2)) Note that the pretty command returns anexpressionina more readable format. Whenyourunthe file, it displays the following result: ans = sin(x) ans = exp(x) ans = x*(log(x) - 1) ans = log(x) ans = (24*cos(5*x))/3125 + (24*x*sin(5*x))/625 - (12*x^2*cos(5*x))/125 + (x^4*cos(5*x))/5 - (4*x^3*sin(5*x))/25 + (x^5*sin(5*x))/5 2 4 24 cos(5 x) 24 x sin(5 x) 12 x cos(5 x) x cos(5 x) ----------- + ------------- - -------------- + ----------- - 3125 625 125 5 3 5
- 3. 4 x sin(5 x) x sin(5 x) ------------- + ----------- 25 5 ans = -1/(4*x^4) ans = tan(x) 2 x (3 x - 5 x + 1) ans = - (7*x^6)/12 - (3*x^5)/5 + (5*x^4)/8 + x^3/2 6 5 4 3 7 x 3 x 5 x x - ---- - ---- + ---- + -- 12 5 8 2 Finding Definite Integral Using MATLAB By definition, definite integralis basically the limit of a sum. We use definite integrals to find areas suchas the area betweena curve and the x-axis and the area betweentwo curves. Definite integrals canalso be used inother situations, where the quantity required canbe expressed as the limit of a sum. The int command canbe used for definite integrationby passing the limits over whichyouwant to calculate the integral. To calculate we write, int(x, a, b) For example, to calculate the value of we write: int(x, 4, 9) MATLAB executes the above statement and returns the following result: ans = 65/2 Following is Octave equivalent of the above calculation: pkg load symbolic symbols x = sym("x");
- 4. f = x; c = [1, 0]; integral = polyint(c); a = polyval(integral, 9) - polyval(integral, 4); display('Area: '), disp(double(a)); Analternative solutioncanbe givenusing quad() functionprovided by Octave as follows: pkg load symbolic symbols f = inline("x"); [a, ierror, nfneval] = quad(f, 4, 9); display('Area: '), disp(double(a)); Example 1 Let us calculate the area enclosed betweenthe x-axis, and the curve y = x3−2x+5 and the ordinates x = 1 and x = 2. The required area is givenby: Create a script file and type the following code: f = x^3 - 2*x +5; a = int(f, 1, 2) display('Area: '), disp(double(a)); Whenyourunthe file, it displays the following result: a = 23/4 Area: 5.7500 Following is Octave equivalent of the above calculation: pkg load symbolic symbols x = sym("x"); f = x^3 - 2*x +5; c = [1, 0, -2, 5]; integral = polyint(c); a = polyval(integral, 2) - polyval(integral, 1); display('Area: '), disp(double(a)); Analternative solutioncanbe givenusing quad() functionprovided by Octave as follows: pkg load symbolic symbols
- 5. x = sym("x"); f = inline("x^3 - 2*x +5"); [a, ierror, nfneval] = quad(f, 1, 2); display('Area: '), disp(double(a)); Example 2 Find the area under the curve: f(x) = x2 cos(x) for −4 ≤ x ≤ 9. Create a script file and write the following code: f = x^2*cos(x); ezplot(f, [-4,9]) a = int(f, -4, 9) disp('Area: '), disp(double(a)); Whenyourunthe file, MATLAB plots the graph: and displays the following result: a = 8*cos(4) + 18*cos(9) + 14*sin(4) + 79*sin(9) Area: 0.3326 Following is Octave equivalent of the above calculation: pkg load symbolic symbols x = sym("x"); f = inline("x^2*cos(x)"); ezplot(f, [-4,9]) print -deps graph.eps [a, ierror, nfneval] = quad(f, -4, 9);