Double integrals 2

[Petrus]

Hello MHB,
I would like to have tips how to solve the x limits for this problem

,
there

Regards,

 

[Klaas van Aarsen]

Hello MHB,
I would like to have tips how to solve the x limits for this problem

,
there

Regards,

Your area D is bounded by the 2 curves $y=x^6$ and $y=x^{1/5}$.
Where do they intersect?
Or put differently, can you solve that set of 2 equations?

 

[Petrus]

Your area D is bounded by the 2 curves $y=x^6$ and $y=x^{1/5}$.
Where do they intersect?
Or put differently, can you solve that set of 2 equations?

hmm \(\displaystyle x^6=x^{1/5} <=> x_1=1 x_2=0\)

 

[Klaas van Aarsen]

hmm \(\displaystyle x^6=x^{1/5} <=> x_1=1 x_2=0\)

Yep.
Those are the x limits in your problem.

 

[Petrus]

Yep.
Those are the x limits in your problem.

Hello I like Serena,
I forgot to say thanks, I solved it!

For anyone who is intrested how to solve it:
\(\displaystyle \int_0^1\int_{x^6}^{x^{\frac{1}{5}}} x^4y^6 \ dydx\)
We can always use Fubini's theorem and take out the constant and integrate the y so we got

\(\displaystyle \int_0^1 x^4 \left[\frac{y^7}{7}\right]_{x^6}^{x^{\frac{1}{5}}}\)
\(\displaystyle \int_0^1x^4\frac{x^{\frac{7}{5}}}{7}-x^4\frac{x^{42}}{7} dx\)
use the rule \(\displaystyle x^m*x^n=x^{n+m}\) and we can simplify to ( we can take out a constant as well):

\(\displaystyle \frac{1}{7}\int_0^1x^{\frac{27}{5}}-x^{46} dx\)
\(\displaystyle \frac{1}{7} \left[ \frac{5x^{\frac{32}{5}}}{32} - \frac{x^{47}}{47} \right]_0^1 = \frac{29}{1504}\)

Regards,