# Finding Area using Integrals

As part of my Calculus 2 class, I have created a practice exam to help my students study.  In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.  I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

## Find the area bounded by the function $$f(x)=8x$$, $$g(x)=\frac{8}{x}$$, and $$h(x)=x^{2}$$.

Drawing a sketch of these graphs, we get the following picture. In order to find the area, we cut this into smaller cross sections, then add up the areas of the cross sections. Here we will cut out rectangles as shown below. Note that the height of the rectangles has a different function value in the first portion than it does in the second portion, so we will split this into two parts. In the first section, we get that
\begin{align*} A_{cs}&=h*w \\ &=(y_{b}-y_{s})\Delta x, \end{align*} where $$y_{b}$$ is the bigger $$y$$ value and $$y_{s}$$ is the smaller $$y$$ value. In this case, we note from the graph that $$y_{b}=8x$$ and $$y_{s}=x^{2}$$. Therefore,
\begin{align*} A_{cs}=(8x-x^{2})\Delta x. \end{align*}

In order to find the total area over this interval, we then find that
\begin{align*} A_{T1}=\int_{a}^{b}(8x-x^{2})dx. \end{align*} Where is $$a$$ is the smallest $$x$$ value and $$b$$ is the largest $$x$$ value where these cross sections work.

We find $$a$$ as the intersection of $$8x$$ and $$x^{2}$$. That is, we solve
\begin{align*} 8x&=x^{2} \\ 0&=x^{2}-8x \\ 0&=x(x-8) \\ x&=0,8. \end{align*} From the picture, we note that the point we are interested for this example is actually $$a=0$$.

We then find $$b$$ as the intersection of $$8x$$ and $$\frac{8}{x}$$, so we get
\begin{align*}
8x&=\frac{8}{x} \\
8x^{2}&=8 \\
x^{2}&=1 \\
x&=\pm 1.
\end{align*}
Again, by looking at the picture, we note that the correct choice is $$b=1$$.

Therefore,
\begin{align*} A_{T1}&=\int_{0}^{1}(8x-x^{2})dx \\ &=4x^{2}-\frac{x^{3}}{3}|_{0}^{1} \\ &=4-\frac{1}{3}-0=\frac{11}{3}. \end{align*}

If we follow the same process for the second area, we note that we will have $$y_{b}=\frac{8}{x}$$ and $$y_{s}=x^{2}$$. Therefore, we will get that
\begin{align*} A_{cs}&=h*w \\ &=(y_{b}-y_{s})\Delta x \\ &=(\frac{8}{x}-x^{2})\Delta x. \end{align*}

Therefore, the total area of the second section will be
\begin{align*} A_{T2}=\int_{b}^{c}(\frac{8}{x}-x^{2})dx. \end{align*}

We have already found that $$b=1$$, so now we need to find $$c$$. Here, this is the intersection of $$\frac{8}{x}$$ and $$x^{2}$$. Therefore, we get that
\begin{align*} \frac{8}{x}&=x^{2} \\ 8&=x^{3} \\ x&=2. \end{align*} Hence, $$c=2$$.

We then get that
\begin{align*} A_{T2}&=\int_{1}^{2}(\frac{8}{x}-x^{2})dx \\ &=8\ln|x|-\frac{x^{3}}{3}|_{1}^{2} \\ &=8\ln(2)-\frac{8}{3}-(8\ln(1)-\frac{1}{3}) \\ &=8\ln(2)-\frac{7}{3}. \end{align*}

Now that we have found both areas, the total area of the region will be given as the sum of the two areas. Therefore, the total area is
\begin{align*} A_{T}&=A_{T1}+A_{T2} \\ &=\frac{11}{3}+8\ln(2)-\frac{7}{3} \\ &=\frac{4}{3}+8\ln(2). \end{align*}

## Other Resources

I have it linked above, but you can find the practice exam here.  You can find the other solutions to the practice exam in the other posts available here.

If you found the post, or the YouTube video helpful, please like the post or the video.  Also remember to follow the blog and subscribe to the YouTube channel.  By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus.  Thank you.

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