MHBSolving Iterated Integral 3 with Antiderivatives

[Petrus]

Hello MHB,
I got stuck on one exercise,

\(\displaystyle \int_0^1\int_y^{e^y}\sqrt{x}dxdy\)
So I antiderivate respect to x and get
\(\displaystyle \Bigl[\frac{2x^{\frac{3}{2}}}{3} \Bigr]_y^{e^y}\)
so we got:

\(\displaystyle \int_0^1\frac{2e^{\frac{2y}{3}}-2y^{\frac{2}{3}}}{3}\)

So I did try antiderivate that but as soon as I try antiderivate \(\displaystyle 2e^{\frac{2y}{3}}\) I would get \(\displaystyle \frac{2e^{\frac{6y}{3}}}{y}\) and we got zero in our limit that means I done something wrong and I can't see what I done wrong
edit: in the limits it's \(\displaystyle e^y\) it does not look clearly

Regards,

 

[topsquark]

Hello MHB,
I got stuck on one exercise,

\(\displaystyle \int_0^1\int_y^{e^y}\sqrt{x}dxdy\)
So I antiderivate respect to x and get
\(\displaystyle \Bigl[\frac{2x^{\frac{3}{2}}}{3} \Bigr]_y^{e^y}\)
so we got:

\(\displaystyle \int_0^1\frac{2e^{\frac{2y}{3}}-2y^{\frac{2}{3}}}{3}\)

So I did try antiderivate that but as soon as I try antiderivate \(\displaystyle 2e^{\frac{2y}{3}}\) I would get \(\displaystyle \frac{2e^{\frac{6y}{3}}}{y}\) and we got zero in our limit that means I done something wrong and I can't see what I done wrong
edit: in the limits it's \(\displaystyle e^y\) it does not look clearly

Regards,

$$\int e^{2y/3} dy = \frac{3}{2}e^{2y/3} + C$$

-Dan

 

[Petrus]

Thanks Dan!
I guess I never notice that I confused myself when I integrated with e but I solved it, here is how I solved it:

\(\displaystyle \int_0^1\int_y^{e^y}\sqrt{x}dxdy\)
So I antiderivate respect to x and get
\(\displaystyle \Bigl[\frac{2x^{\frac{3}{2}}}{3} \Bigr]_y^{e^y}\)
so we got:

\(\displaystyle \int_0^1\frac{2e^{\frac{2y}{3}}-2y^{\frac{2}{3}}}{3}dy\)
\(\displaystyle \frac{1}{3} \)\(\displaystyle \Bigl[ 3e^{\frac{2y}{3}}-\frac{6y^{\frac{5}{3}}}{5} \Bigr]_0^1= e^{\frac{2}{3}}-\frac{7}{5}\)

Regards,

 

[MarkFL]

Hello Petrus,

I agree that:

\(\displaystyle \int_y^{e^y}\sqrt{x}\,dx=\frac{2}{3}\left[x^{\frac{3}{2}} \right]_y^{e^y}\)

However, for the next step, I would write:

\(\displaystyle \frac{2}{3}\int_0^1 e^{\frac{3y}{2}}-y^{\frac{3}{2}}\,dy=\frac{2}{3}\left[\frac{2}{3}e^{\frac{3y}{2}}-\frac{2}{5}y^{\frac{5}{2}} \right]_0^1\)

Do you see where you went wrong with the exponents? Can you finish now?

 

[Petrus]

Hello Petrus,

I agree that:

\(\displaystyle \int_y^{e^y}\sqrt{x}\,dx=\frac{2}{3}\left[x^{\frac{3}{2}} \right]_y^{e^y}\)

However, for the next step, I would write:

\(\displaystyle \frac{2}{3}\int_0^1 e^{\frac{3y}{2}}-y^{\frac{3}{2}}\,dy=\frac{2}{3}\left[\frac{2}{3}e^{\frac{3y}{2}}-\frac{2}{5}y^{\frac{5}{2}} \right]_0^1\)

Do you see where you went wrong with the exponents? Can you finish now?

Hello Mark,
I want to thank you for taking your time and checking my soloution!

Do you see where you went wrong with the exponents? Can you finish now?

Now I see what I do misstake. I accident confused myself and did write

\(\displaystyle \int_0^1\frac{2e^{\frac{2y}{3}}-2y^{\frac{2}{3}}}
{3}dy\)

while it should be

\(\displaystyle \int_0^1\frac{2e^{\frac{3y}{2}}-2y^{\frac{3}{2}}}{3}dy\)

Can you finish now?

Yes I can, I get now the answer \(\displaystyle \frac{4}{9}e^{\frac{3}{2}}-\frac{32}{45}\)

Regards,