MHB 232.15.3.50 Reverse the order of integration in the following integral

[karush]

$\textsf{Reverse the order of integration in the following integral }$
\begin{align*}\displaystyle
I&=\int_0^1 \int_2^{2e^x}f(x,y) \quad dy \, dx
\end{align*}
$\textit{From the integral we have that}$
$$0\leq x\leq 1 \quad \textit{and} \quad 2\leq y\leq 2e^x$$
$\textit{So, we get that}$
$$y\leq 2e^x \Rightarrow \frac{y}{2}\leq e^x \Rightarrow \ln \left (\frac{y}{2}\right )\leq x$$
$\textit{Therefore, we get that}$
$$\ln \left (\frac{y}{2}\right )\leq x\leq 1 \quad
\textit{and} \quad 2\leq y\leq 2e^x\leq 2e^1=2e$$
$\textit{So, by changing the order of integrals we get the following}$
$$I=\int_2^{2e} \int_{\ln \left (\frac{y}{2}\right )}^1f(x,y) \quad dx \, dy $$
Ok just see if this is ok
presume this is as far as we can go.

 

[tkhunny]

I'm not going to answer your question. I will offer the following comments:

1) Part of the learning of mathematics is acquiring confidence in your own work.

2) Confirmation from someone else will never be as valuable to you as your own style that allows you to verify your own work and be confident in your result even if someone else disagrees with you.

3) I cannot imagine your fear when you hand in an exam and you have no idea if you did well. I cannot imagine having to wait for a grade to have any confidence in your work.

4) "as far as we can go" - This is very odd. It is also important to know when you are done. Why would you go any farther if you are done?

You do a lot of good work. In my opinion, you don't need nearly as much confirmation as you seem to think. Don't take this to mean that you should stop submitting problems for confirmation. If you need help, you should get it. From my experience, you're rewarding to have around because you actually interact, and ask good, searching questions, and appear actually to be learning. Many do those thing, but many don't.

My views. I welcome others'.

 

[karush]

Well I am retired so I am mostly on my own

if were not for this forum I could not keep going with it

I do audit classes but have physical handicaps that make that hard
I don't get much out of it. I was in one today couldn't hear anything.

if it were not for this forum I would just be a zombie

 

[tkhunny]

Well I am retired so I am mostly on my own

if were not for this forum I could not keep going with it

I do audit classes but have physical handicaps that make that hard
I don't get much out of it. I was in one today couldn't hear anything.

if it were not for this forum I would just be a zombie

I hear ya. Keep learning all the time. I've always accused myself of being quite able to sit in a corner and rot. Close to 20 years ago, I began to wonder if I still could learn! I had become lazy and I had forgotten much. I grabbed every math book I ever used - high school and beyond - to see if I could reverse the trend. Since then, I've answered some 70,000 questions on various math help web sites. I guess that suggests that I did manage to reverse the trend. You are an inspiration, I'm not suggesting any big change for you, just maybe a slightly different presentation. Perhaps, "I think I got this one." or "I fixed my error and now I have it!" instead of "Is that right?" or "How close am I?" :-)

 

[Prove It]

$\textsf{Reverse the order of integration in the following integral }$
\begin{align*}\displaystyle
I&=\int_0^1 \int_2^{2e^x}f(x,y) \quad dy \, dx
\end{align*}
$\textit{From the integral we have that}$
$$0\leq x\leq 1 \quad \textit{and} \quad 2\leq y\leq 2e^x$$
$\textit{So, we get that}$
$$y\leq 2e^x \Rightarrow \frac{y}{2}\leq e^x \Rightarrow \ln \left (\frac{y}{2}\right )\leq x$$
$\textit{Therefore, we get that}$
$$\ln \left (\frac{y}{2}\right )\leq x\leq 1 \quad
\textit{and} \quad 2\leq y\leq 2e^x\leq 2e^1=2e$$
$\textit{So, by changing the order of integrals we get the following}$
$$I=\int_2^{2e} \int_{\ln \left (\frac{y}{2}\right )}^1f(x,y) \quad dx \, dy $$
Ok just see if this is ok
presume this is as far as we can go.

With vertical strips, each strip is bounded on the bottom by $\displaystyle \begin{align*} y = 2 \end{align*}$ and on the top by the function $\displaystyle \begin{align*} y = 2\,\mathrm{e}^x \end{align*}$, with the strips summed up over $\displaystyle \begin{align*} x = 0 \end{align*}$ and $\displaystyle \begin{align*} x = 1 \end{align*}$.

The corner points are $\displaystyle \begin{align*} \left( 0 , 2 \right) \end{align*}$ and $\displaystyle \begin{align*} \left( 1 , 2\,\mathrm{e} \right) \end{align*}$.

If instead we use horizontal strips, then they are bounded on the left by $\displaystyle \begin{align*} x = \ln{ \left( \frac{y}{2} \right) } \end{align*}$ and on the right by $\displaystyle \begin{align*} x = 1 \end{align*}$, with the strips summed up over $\displaystyle \begin{align*} y = 2 \end{align*}$ and $\displaystyle \begin{align*} y = 2\,\mathrm{e} \end{align*}$.

So that means with the order of integration reversed, we have

$\displaystyle \begin{align*} \int_0^1{\int_2^{2\,\mathrm{e}^x}{f\left( x, y \right) \,\mathrm{d}y}\,\mathrm{d}x} = \int_2^{2\,\mathrm{e}}{\int_{\ln{ \left( \frac{y}{2} \right) }}^1{f\left( x, y \right) \,\mathrm{d}x }\,\mathrm{d}y} \end{align*}$