1 ≤x≤4
se requiere:
I = \int_1^4 x ln(x) dxliteral a. Usando el método de Cuadratura de Gauss con 2 términos
x_a = \frac{b+a}{2} + \frac{b-a}{2}x_0 = \frac{4+1}{2} + \frac{4-1}{2}\Big(\frac{-1}{\sqrt{3}} \Big)xa =1.6339745962155612
x_b = \frac{b+a}{2} + \frac{b-a}{2}x_1 = \frac{4+1}{2} + \frac{4-1}{2}\Big(\frac{1}{\sqrt{3}} \Big)xb =3.366025403784439
I \cong \frac{b-a}{2}(f(x_a) + f(x_b)) I \cong \frac{4-1}{2}(x_a ln(x_a) + x_b ln(x_b))I = 7.33164251999249
literal b. De la fórmula , despejar el valor del error<0.0001
\Big|\frac{(b-a)}{180}h^4 f^{(4)} (\xi)\Big| <0.0001; \xi \in[a,b] h^4 <0.0001\frac{180}{(4-1)}\frac{1}{f^{(4)} (\xi)} h^4 < 0.006\frac{1}{f^{(4)} (\xi)} h <\Big(0.006\frac{1}{f^{(4)} (\xi)}\Big)^{1/4}obteniendo la 4ta derivada de la función:
f(x) = x ln(x) f'(x) = ln(x) + x\Big(\frac{1}{x} \Big) = ln(x) +1 f''(x) = \frac{1}{x} f'''(x) = -\frac{1}{x^2} f^{(4)}(x) = 2\frac{1}{x^3}se tiene que:
h <\Big(0.006\frac{1}{f^{(4)} (\xi)}\Big)^{1/4} h <\Big(0.006\frac{1}{2\frac{1}{\xi^3}}\Big)^{1/4} h <\Big(0.003\xi^3\Big)^{1/4} h <(0.003)^{1/4}\xi^{3/4}en el peor de los csis, se toma el valor menor de ξ =1
h <(0.003)^{1/4} h<0.2340347319320716