How do I find the area of the region bounded by following?

Drioton
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Homework: Misplaced Thread -- Member warned to post homework questions in the appropriate area
Using integrals, consider the 7 requirements:
Any my attempted solution that I have no idea where I am going:
And the other one provides the graph:
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I'm not sure what you did in your approach besides rewriting the equations and it is difficult to read or understand.

This shouldn't be the first "find the area" problem you encounter. How did you solve the previous problems?

You have marked two intersections of the different equations already. What are their coordinates? Where is a third intersection at the boundary of your area?

You can integrate over x or over y. How would you set up the integrals? Which one looks easier?
 
mfb said:
I'm not sure what you did in your approach besides rewriting the equations and it is difficult to read or understand.

This shouldn't be the first "find the area" problem you encounter. How did you solve the previous problems?

You have marked two intersections of the different equations already. What are their coordinates? Where is a third intersection at the boundary of your area?

You can integrate over x or over y. How would you set up the integrals? Which one looks easier?
F07813A0-8387-4907-89FA-876E87CFD491.jpeg

Here, but these two problems are different. they only consist of one function. Whereas this question has “x is greater than or equal to zero” (what is the meaning of this with respect to the problem) part, and has two functions. Besides the solution in yellow paper, it is unclear how I figure these out?
 
You could rotate your images in the correct orientation, that would help already.
Drioton said:
Here, but these two problems are different. they only consist of one function.
It is still the area between four boundary lines. Here you just have three.

In one of the two options for the integration it is advisable to split the area into two regions, calculate their area separately and then add them. In the other case (the easier one!) this is not necessary.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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