MHB Double Integrals 2: Solving X Limits

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The discussion focuses on solving the x limits for the area bounded by the curves y=x^6 and y=x^{1/5}. The intersection points of these curves are identified as x=0 and x=1. A double integral is set up to calculate the area, expressed as ∫_0^1∫_{x^6}^{x^{1/5}} x^4y^6 dy dx. The integration process involves applying Fubini's theorem and simplifying the resulting expressions. The final result of the integral is calculated to be 29/1504.
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Hello MHB,
I would like to have tips how to solve the x limits for this problem
b867adbf43d51b76f0ac0beaa6f58a1.png
,
there
63f86405842dc75e05f46ff22f62e41.png
Regards,
 
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Petrus said:
Hello MHB,
I would like to have tips how to solve the x limits for this problem
b867adbf43d51b76f0ac0beaa6f58a1.png
,
there
63f86405842dc75e05f46ff22f62e41.png
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?
 
I like Serena said:
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 $$x^6=x^{1/5} <=> x_1=1 x_2=0$$
 
Petrus said:
hmm $$x^6=x^{1/5} <=> x_1=1 x_2=0$$

Yep.
Those are the x limits in your problem.
 
I like Serena said:
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 interested how to solve it:
$$\int_0^1\int_{x^6}^{x^{\frac{1}{5}}} x^4y^6 \ dydx$$
We can always use Fubini's theorem and take out the constant and integrate the y so we got

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

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

Regards,
 
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