15.2.66 Solve by reversing the reversing the order of integration

[karush]

solve by reversing the reversing the order of integration
this was given:
\begin{align*}\displaystyle
I&=\int_0^8 \int_{\sqrt[3]{x}}^2
\left[\frac{x}{y^7+1}\right]dy \, dx\\
\end{align*}
ok I put this in an dbl int calculor but it turned the order around to
\begin{align*}\displaystyle
I&=\int_{\sqrt[3]{x}}^2 \int_0^8
\left[\frac{x}{y^7+1}\right]dy \, dx\\
\end{align*}

 

[Greg]

Re: 15.2.66 solve by reversing the reversing the order of integration

So what's your question? :)

 

[MarkFL]

Re: 15.2.66 solve by reversing the reversing the order of integration

solve by reversing the reversing the order of integration
this was given:
\begin{align*}\displaystyle
I&=\int_0^8 \int_{\sqrt[3]{x}}^2
\left[\frac{x}{y^7+1}\right]dy \, dx\\
\end{align*}

I would begin by plotting the region over which we are to integrate:

View attachment 7254

The integral as given is using vertical strips...and we need to use horizontal strips instead.

In order to reverse the order of integration, we need to observe that we have:

\(\displaystyle 0\le x\le y^3\)...the upper limit was obtained from $y=\sqrt[3]{x}$...

\(\displaystyle 0\le y\le 2\)

And so we have:

\(\displaystyle I=\int_0^2\int_0^{y^3}\frac{x}{y^7+1}\,dx\,dy\)

Now you will be able to integrate...can you continue?

 

[karush]

Re: 15.2.66 solve by reversing the reversing the order of integration

we got
$\frac{\ln\left({129}\right)}{14}$

 

[MarkFL]

Re: 15.2.66 solve by reversing the reversing the order of integration

we got
$\frac{\ln\left({129}\right)}{14}$

Yes:

\(\displaystyle I=\int_0^2\int_0^{y^3}\frac{x}{y^7+1}\,dx\,dy=\int_0^2\frac{1}{y^7+1}\int_0^{y^3}x\,dx\,dy=\frac{1}{2}\int_0^2\frac{1}{y^7+1}\int_0^{y^3}\left[x^2\right]_0^{y^3}\,dy\)

\(\displaystyle I=\frac{1}{2}\int_0^2\frac{y^6}{y^7+1}\,dy=\frac{1}{14}\int_0^2\frac{7y^6}{y^7+1}\,dy=\frac{1}{14}\int_1^{129}\frac{1}{u}\,du=\frac{1}{14}\left[\ln(u)\right]_1^{129}=\frac{1}{14}\ln(129)\)