Series Solutions Problems
Test for Divergence

Example 1.

We are given the integral

(1)   \begin{equation*}\int _{0}^{3}\int _{0}^{1}2xy^{2}dxdy \end{equation*}

Note that we are integrating the function z =2xy over the region bounded by

(2)   \begin{equation*}x =0 ,x =1 ,y =0 ,y =3 \end{equation*}

Integrating with respect to x , the value of y remain constant that is

(3)   \begin{align*}\int _{0}^{3}\int _{0}^{1}2xy^{2}dxdy =2\int _{0}^{3}y^{2}\left [\int _{0}^{1}xdx\right ]dy \\ =2\int _{0}^{3}y^{2}\left [\frac{1}{2}x^{2}\right ]_{0}^{1}dy \\ =\left [\frac{1}{3}y^{3}\right ]_{0}^{3} =9\end{align*}

Example 2.

We are given

(4)   \begin{equation*}\int _{0}^{1}\int _{x^{2}}^{\sqrt{x}}dydx \end{equation*}

We are integrating the function z =1 over the region bounded by

(5)   \begin{equation*}x =0 ,x =1 ,y =x^{2} ,y =\sqrt{x} \end{equation*}

This integral provides the volume of a solid with the above region as base and height 1. With the same token, the value of this integral provides the area of the bounded region (why?) Therefore we write

(6)   \begin{align*}\int _{0}^{1}\int _{x^{2}}^{\sqrt{x}}dydx =\int _{0}^{1}\left [y\right ]_{x^{2}}^{\sqrt{x}}dx \\ =\int _{0}^{1}\left (\sqrt{x} -x^{2}\right )dx \\ =\left [\frac{2}{3}x^{3/2} -\frac{1}{3}x^{3}\right ]_{0}^{1} \\ =\frac{1}{3}\end{align*}

Example 3.

We want to integrate the function z =\ln y over the region bounded by y =2 ,y =3,x =0 and x =\frac{1}{y}. Note that we write the integral in two steps. First, we determine the given region between two constant value (like a belt) with respect to either variables when possible. Here, we showed them with red arrows. Then inside the belt portion, we determine the beginning and ending values with respect to the other variable. In the graph, we showed them with green arrows.

Now we evaluate the integral

(7)   \begin{align*}\int _{2}^{3}\int _{0}^{\frac{1}{y}}\ln ydxdy =\int _{2}^{3}\ln y\left [x\right ]_{0}^{1/y}dy \\ =\int _{2}^{3}\frac{\ln y}{y}dy \\ =\frac{1}{2} \left [ (\ln {3})^2 - (\ln {2})^2 \right ] \end{align*}

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