Find n for the shaded region

1. May 16, 2009

Emethyst

1. The problem statement, all variables and given/known data
Find the value of n so that the area of the shaded region (refer to attached picture) in the following diagram is a) 50% of the area of the unit square b) 80% of the area of the unit square

2. Relevant equations
Definate integral properities, fundamental theorem of calculus

3. The attempt at a solution
My teacher says this is an easy question, but I cannot seem to solve it. My first guess was to take the integral of the two functions and then use the percentages as the answers for the integral, solving for n in each case (in this case I made 50% = 1/2 and 80% = 4/5). This did not get me very far for I came up with square roots and fractions in my answers when they should be simply be one number answers. Am I doing this question right through this method, or is this the wrong procedure and do I need to do something different to find the correct answers? Any help would be very welcome, thanks in advance. PS: sorry for the bad picture :tongue:

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2. May 17, 2009

Staff: Mentor

The picture has been pending approval for several hours. Can you describe the region in words?

3. May 17, 2009

Emethyst

There are two functions, y = x^n and y = x^1/n located in region I of the graph. The area to be found is the area in-between these two functions, and runs from their point of intersection at (0,0) to the other point at (1,1). y = x^n is located below y = x^1/n (though i'm sure you probably already figured that out :tongue:).

4. May 17, 2009

squidsoft

$$A=\int_0^1\int_{x^n}^{x^{1/n}} dydx$$
(depending if n>1 or n<1)

Evaluate the integral then solve for n. I get:

$$\frac{n}{n+1}-\frac{1}{n+1}=A$$

Last edited: May 17, 2009
5. May 17, 2009

Emethyst

I understand the last part, but I don't know how to get (n-1)/(n+1) from that integral you used. Unless you mean finding the integral of x^1/n - x^n over [0,1]?

6. May 18, 2009

squidsoft

Hi. Didn't notice you replied until now.

<< solution deleted by berkeman >>

Last edited by a moderator: May 18, 2009
7. May 18, 2009

Emethyst

Thanks squidsoft, I managed to solve that one.