Método del punto medio

El procedimiento numérico que vamos a utilizar para obtener la integral definida va a ser el del 'punto medio'.

Vamos a aproximar la integral definida ${\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}$, sumando las áreas de cada uno de los rectángulos.

El área del rectángulo de la figura es f(x_m)(x_i+1-x_i)= f(x_m)Δx_i, donde x_m es el punto medio

Para ello definimos la función Integral

function suma = Integral( f, x )
    suma=0;
    for  i=1:length(x)-1
        xm=(x(i+1)+x(i))/2;
        suma=suma+f(xm)*(x(i+1)-x(i));
    end
end

Ejemplos

• ${\displaystyle {{\displaystyle \underset{0}{\overset{1}{\int }}xdx=\frac{x^2}{2}|}}_0^1=\frac{1}{2}}$

>> f=@(x) x;
>> x=linspace(0,1,30);
>> Integral(f,x)
ans =    0.5000

Alternativamente

>> x=0:0.01:1;
>> suma=Integral(f,x)
suma =    0.5000

• ${\displaystyle {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{sen}xdx={-\cos x|}_0^{\frac{\pi }{2}}}=1}$

>> f=@(x) sin(x);
>> x=linspace(0,pi/2,100);
>> Integral(f,x)
ans =   1.0000

Alternativamente

>> f=@(x) sin(x);
>> x=0:0.1:pi/2;
>> format long
>> suma=Integral(f,x)
suma =   0.929650104125955

>> x=0:0.0001:pi/2;
>> suma=Integral(f,x)
suma =   0.999903673621882

>> x=0:1e-6:pi/2;
>> suma=Integral(f,x)
suma =   0.999999673205143

>> x=0:1e-8:pi/2;
>> x=0:1e-8:pi/2;
suma=Integral(f,x)
suma =   0.999999993205169

Integración numérica con funciones MATLAB

La función integral nos da el valor aproximado de la integral cuando le introducimos la función a integrar y los límites de integración.

q=integral(function, a, b);

NOTA.- L a x funciona como vector en integral no olvidar el (.)

• ${\displaystyle {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\mathrm{sen}xdx={-\cos x|}_0^{\frac{\pi }{2}}}=1}$

>> f=@(x) sin(x);
>> q=integral(f,0,pi/2)
q =    1.0000

• ${\displaystyle {\displaystyle \underset{-5}{\overset{5}{\int }}\sqrt{25-x^2}dx}=\frac{25}{2}\pi }$

>> f=@(x) sqrt(25-x.^2); %Importante el .
>> q=integral(f,-5,5)
q =   39.2699

Con la calculadora, 25π/2=39,26990817

Ejercicios

1. ${\displaystyle {\displaystyle \underset{1}{\overset{2}{\int }}(x^2-2x+3)dx}}$

>> f=@(x) x.^2-2*x+3;
>> q=integral(f,1,2)
q =    2.3333

Con el procedimiento del punto medio

>> f=@(x) x^2-2*x+3;
>> x=linspace(1,2,200);
>> Integral(f,x)
ans =    2.3333

2. ${\displaystyle {\displaystyle \underset{0}{\overset{8}{\int }}(\sqrt{2x}+\sqrt[3]{x})dx}}$

>> f=@(x) sqrt(2*x)+nthroot(x,3);
>> q=integral(f,0,8)
q =   33.3333

Con el procedimiento del punto medio

>> f=@(x) sqrt(2*x)+nthroot(x,3);
>> x=linspace(0,8,300);
>> Integral(f,x)
ans =   33.3342

>> x=linspace(0,8,3000);
>> Integral(f,x)
ans =   33.3334

>> x=linspace(0,8,30000);
>> Integral(f,x)
ans =   33.3333

>> x=0:0.0001:8;
>> suma=Integral(f,x)
suma =   33.3333

3. ${\displaystyle {\displaystyle \underset{1}{\overset{4}{\int }}\frac{1+\sqrt{x}}{x^2}dx}}$

>> f=@(x) (1+sqrt(x))./x.^2;
>> q=integral(f,1,4)
q =   1.7500

Con el procedimiento del punto medio

>> f=@(x) (1+sqrt(x))/x^2;
>> x=linspace(1,4,1000);
>> Integral(f,x)
ans =    1.7500

>> f=@(x) (1+sqrt(x))/x^2;
>> x=1:0.0001:4;
>> suma=Integral(f,x)
suma =    1.7500

4. ${\displaystyle {\displaystyle \underset{2}{\overset{6}{\int }}\sqrt{x-2}dx}}$

>> f=@(x) sqrt(x-2);
>> q=integral(f,2,6)
q =    5.3333

Con el procedimiento del punto medio

>> f=@(x) sqrt(x-2);
>> x=linspace(2,6,1000);
>> Integral(f,x)
ans =    5.3333

5. ${\displaystyle {\displaystyle \underset{0}{\overset{-3}{\int }}\frac{dx}{\sqrt{25+3x}}}}$

>> f=@(x) 1./ sqrt(25+3*x);
>> q=integral(f,0,-3)
q =   -0.6667

Con el procedimiento del punto medio

>> f=@(x) 1/ sqrt(25+3*x);
>> x=linspace(0,-3,1000);
>> Integral(f,x)
ans =     -0.6667

6. ${\displaystyle {\displaystyle \underset{-2}{\overset{-3}{\int }}\frac{dx}{x^2-1}}}$

>> f=@(x) 1./ (x.^2-1);
>> q=integral(f,-2,-3)
q =   -0.2027

Con el procedimiento del punto medio

>> f=@(x) 1/ (x^2-1);
>> x=linspace(-2,-3,1000);
>> Integral(f,x)
ans =   -0.2027

7. ${\displaystyle {\displaystyle \underset{0}{\overset{1}{\int }}\frac{xdx}{x^2+3x+2}}}$

>> f=@(x) x./(x.^2+3*x+2);
>> q=integral(f,0,1)
q =    0.1178

Con el procedimiento del punto medio

>> f=@(x) x/(x^2+3*x+2);
>> x=linspace(0,1,1000);
>> Integral(f,x)
ans =    0.1178

8. ${\displaystyle {\displaystyle \underset{-1}{\overset{1}{\int }}\frac{x^5}{x+2}dx}}$

>> f=@(x) x.^5./(x+2);
>> q=integral(f,-1,1)
q =   -0.0889

Con el procedimiento del punto medio

>> f=@(x) x^5/(x+2);
>> x=linspace(-1,1,1000);
>> Integral(f,x)
ans =   -0.0889

9. ${\displaystyle {\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{x^2+4x+5}}}$

>> f=@(x) (x.^2+4*x+5).^(-1);
>> a=integral(f,0,1)
a =    0.1419

Con el procedimiento del punto medio

>> x=linspace(0,1,200);
>> suma=Integral(f,x)
suma =    0.1419

10. ${\displaystyle {\displaystyle \underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{co{s}^{2}xdx}}}$

>> f=@(x) 1./(cos(x)).^2;
>> q=integral(f,pi/6,pi/4)
q =    0.4226

Con el procedimiento del punto medio

>> f=@(x) 1/(cos(x))^2;
>> x=linspace(pi/6,pi/4,1000);
>> Integral(f,x)
ans =    0.4226

11. ${\displaystyle {\displaystyle \underset{1}{\overset{e}{\int }}\frac{\mathrm{sen}(\ln x)}{x}dx}}$

>> f=@(x) sin(log(x))./x;
>> a=integral(f,1,exp(1))
a =    0.4597

Con el procedimiento del punto medio

>> x=linspace(1,exp(1),100);
>> suma=Integral(f,x)
suma =    0.4597

Integrales indefinidas con simbólico

1. ${\displaystyle {\displaystyle \int \frac{\sqrt[6]{13+lnx}}{x}dx}}$

>> syms x
>> y=((13+log(x))^(1/6))/x;
>> a=int(y)
a = (6*(log(x) + 13)^(7/6))/7
>> b=diff(a) 
b =(log(x) + 13)^(1/6)/x

2. ${\displaystyle {\displaystyle \int \frac{dx}{(x^2-4x+3)(x+5)}}}$

>> y=(x^2-4*x+3)^-1*(x+5)^-1;
>> z=int(y)
z =log(x - 3)/16 - log(x - 1)/12 + log(x + 5)/48.
>> diff(z)
ans =1/(16*(x - 3)) - 1/(12*(x - 1)) + 1/(48*(x + 5)) 
>> simplify(ans) 
ans =1/((x - 1)*(x - 3)*(x + 5))

3. ${\displaystyle {\displaystyle \int \frac{dx}{(x-1)(x-2)(x+4)}}}$

>> y=(x-1)^-1*(x-2)^-1*(x+4)^-1;
>> z=int(y)
z =log(x - 2)/6 - log(x - 1)/5 + log(x + 4)/3
>> y=diff(z) 
y = 1/(6*(x - 2)) - 1/(5*(x - 1)) + 1/(30*(x + 4))
>> simplify(y)
ans =1/((x - 1)*(x - 2)*(x + 4))

4. ${\displaystyle {\displaystyle \int \frac{5x^4}{x^3-5x^2+4x}dx}}$

>> y=5*x^4/(x^3-5*x^2+4*x);
z=int(y) 
z = 25*x - (5*log(x - 1))/3 + (320*log(x - 4))/3 + (5*x^2)/2
>> y=diff(z)
y =5*x - 5/(3*(x - 1)) + 320/(3*(x - 4)) + 25
>> simplify(y) 
ans = (5*x^3)/(x^2 - 5*x + 4)

5. ${\displaystyle {\displaystyle \int \frac{2x^2+1}{x^3-6x^2+11x-6}}dx}$

>> y=(2*x^2+1)/(x^3-6*x^2+11*x-6);
>> z=int(y)
z = (3*log(x - 1))/2 - 9*log(x - 2) + (19*log(x - 3))/2
>> y=diff(z) 
y =3/(2*(x - 1)) - 9/(x - 2) + 19/(2*(x - 3))

6. ${\displaystyle {\displaystyle \int \frac{dx}{3x^2-x+1}}}$

>> y=(3*x^2-x+1)^-1;
>> z=int(y)
z = (2*11^(1/2)*atan((11^(1/2)*(6*x - 1))/11))/11 
>> diff(z) 
ans =12/(11*((6*x - 1)^2/11 + 1))

7. ${\displaystyle {\displaystyle \int \frac{xdx}{x^2-7x+13}}}$

>> y=x/(x^2-7*x+13);
>> z=int(y)
z = log(x^2 - 7*x + 13)/2 + (7*3^(1/2)*atan((2*3^(1/2)*x)/3 - (7*3^(1/2))/3))/3
>> simplify(z)
ans =log(x^2 - 7*x + 13)/2 + (7*3^(1/2)*atan(3^(1/2)*((2*x)/3 - 7/3)))/3 
>> y=diff(ans) 
y =(2*x - 7)/(2*(x^2 - 7*x + 13)) + 14/(3*(3*((2*x)/3 - 7/3)^2 + 1))
>> simplify(y) 
ans = x/(x^2 - 7*x + 13)

8. ${\displaystyle {\displaystyle \int \frac{x^4}{x^4-1}}dx}$

>> y=x^4/(x^4-1);
>> z=int(y)
z = x - atan(x)/2 - atanh(x)/2
>> y=diff(z)
y =1/(2*(x^2 - 1)) - 1/(2*(x^2 + 1)) + 1
>> simplify(y) 
ans =x^4/(x^4 - 1)

9. ${\displaystyle {\displaystyle \int \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)}}dx}$

>> y=(x^3+x^2+x+3)/((x^2+1)*(x^2+3));
>> z=int(y) 
z = log(x^2 + 3)/2 + atan(x)
>> y=diff(z)
y =x/(x^2 + 3) + 1/(x^2 + 1)

10. ${\displaystyle {\displaystyle \int \frac{dx}{x^3+1}}}$

>> y=1/(x^3+1);
>> z=int(y) 
z =log(x + 1)/3 - log((x - 1/2)^2 + 3/4)/6 + 
(3^(1/2)*atan((2*3^(1/2)*(x - 1/2))/3))/3 
>> y=diff(z)
y = 1/(3*(x + 1)) + 2/(3*((4*(x - 1/2)^2)/3 + 1)) - 
(2*x - 1)/(6*((x - 1/2)^2 + 3/4))
>> simplify(y) 
ans = 1/(x^3 + 1)

11. ${\displaystyle {\displaystyle \int {\mathrm{sen}}^{4}xdx}}$

>> y=(sin(x))^4;
>> z=int(y) 
z =(3*x)/8 - sin(2*x)/4 + sin(4*x)/32 
>> y=diff(z) 
y = cos(4*x)/8 - cos(2*x)/2 + 3/8 
>> simplify(y) 
ans = (cos(2*x) - 1)^2/4 
>> expand(ans) 
ans = cos(x)^4 - 2*cos(x)^2 + 1
>> simplify(ans)
ans =(cos(x)^2 - 1)^2 =(-sin(x)^2)^2=sin(x)^4

12. ${\displaystyle {\displaystyle \int \sqrt{1-x^2}dx}}$

>> y=sqrt(1-x^2);
>> z=int(y)
z = asin(x)/2 + (x*(1 - x^2)^(1/2))/2
>> y=diff(z)
y = 1/(2*(1 - x^2)^(1/2)) - x^2/(2*(1 - x^2)^(1/2)) + (1 - x^2)^(1/2)/2
>> simplify(y) 
ans = (1 - x^2)^(1/2)

13. ${\displaystyle {\displaystyle \int \sqrt{x^2+6x-7}dx}}$

>> y=sqrt(x^2+6*x-7);
>> z=int(y)
z =(x/2 + 3/2)*(x^2 + 6*x - 7)^(1/2) - 8*log(x + (x^2 + 6*x - 7)^(1/2) + 3)
>> y=diff(z) 
y =(x^2 + 6*x - 7)^(1/2)/2 - (8*((2*x + 6)/(2*(x^2 + 6*x - 7)^(1/2)) + 1))
/(x + (x^2 + 6*x - 7)^(1/2) + 3) + ((2*x + 6)*(x/2 + 3/2))
/(2*(x^2 + 6*x - 7)^(1/2))
>> simplify(y) 
ans =(x^2 + 6*x - 7)^(1/2)

14. ${\displaystyle {\displaystyle \int }x·\mathrm{arctg}\left(\frac{x^2+9}{9}\right)dx}$

>> y=x*atan((x^2+9)/9);
>> z=int(y) 
z = (9*atan(x^2/9 + 1))/2 - (9*log(x^4 + 18*x^2 + 162))/4 + 
(x^2*atan(x^2/9 + 1))/2 
>> y=diff(z) 
y = x/((x^2/9 + 1)^2 + 1) - (9*(4*x^3 + 36*x))/(4*(x^4 + 18*x^2 + 162)) 
+ x*atan(x^2/9 + 1) + x^3/(9*((x^2/9 + 1)^2 + 1))
>> simplify(y) 
ans = x*atan(x^2/9 + 1)