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EXAMPLE 5 Evaluate the iterated integral 5 0 5 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating sin(y2) dy. But it's impossible to do so in finite terms since sin(y2) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have 5 0 5 x sin(y2) dy dx

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LammettHash

[tex]\displaystyle\int_0^5\int_x^5\sin(y^2)\,\mathrm dy\,\mathrm dx[/tex]

The integration region is the triangle in the [tex]x,y[/tex] plane bounded by the lines [tex]y=5[/tex], [tex]y=x[/tex], and [tex]x=0[/tex]. Reversing the order of integration, this is equal to

[tex]\displaystyle\int_0^5\int_0^y\sin(y^2)\,\mathrm dx\,\mathrm dy=\int_0^5y\sin(y^2)\,\mathrm dy[/tex]

For the remaining integral, let [tex]u=y^2\implies\mathrm du=2y\,\mathrm dy[/tex]:

[tex]\displaystyle\int_0^5y\sin(y^2)\,\mathrm dy=\frac12\int_0^{25}\sin u\,\mathrm du=-\frac12\cos u\bigg|_0^{25}=\frac{1-\cos(25)}2[/tex]

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