Respuesta :

First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

[tex]x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx[/tex]
[tex]x=1\implies t=\dfrac{2-6}4=-1[/tex]
[tex]x=5\implies t=\dfrac{10-6}4=1[/tex]

So,

[tex]\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt[/tex]

Let [tex]f(t)=2\sin(2t+3)^2[/tex]. With [tex]n=4[/tex], we're looking for coefficients [tex]c_i[/tex] and nodes [tex]x_i[/tex], with [tex]1\le i\le4[/tex], such that

[tex]\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)[/tex]

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

[tex]\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)[/tex]
[tex]\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790[/tex]
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