# Definite Integral Problem

1. Jan 20, 2007

### Dorothy Weglend

1. The problem statement, all variables and given/known data
By appealing to geometric evidence show that
$$\int_0^8x^n\,dx + \int_0^1 x^{1/n}\,dx = 1$$
for n a positive integer.

2. Relevant equations
Fundamental theorem of calculus, power rule for integration.

3. The attempt at a solution

I integrated. For the first integral, I get:

$$\frac{8^{n+1}}{n+1}$$

and for the second:

$$\frac{n}{n+1}$$

As an experiment, I tried this for a few values of n, for example, n=1 gives $8^2/2 + 1/2 = 32.5 [/tex], which is certainly not 1. So obviously something is awry here. I think I integrated properly, but perhaps not. Can someone shed some light on this for me? Thank you, Dorothy Last edited: Jan 20, 2007 2. Jan 20, 2007 ### arildno The "8" should be a "1". 3. Jan 20, 2007 ### HallsofIvy Staff Emeritus Arildno's point is that $$\int_0^1x^n\,dx + \int_0^1 x^{1/n}\,dx = \frac{1}{n+1}+ \frac{n}{n+1}= 1$$ But what you give is NOT always equal to 1. Since the problem says "by appealing to geometric evidence", those integrals are the areas of what regions? 4. Jan 20, 2007 ### Dorothy Weglend Thank you both, so much! It must be a mistake in the text. Now the problem seems solvable. To answer your question, it seems to me that the geometric argument would go something like this: the curve x^n and x^1/n are mirror images of each other, reflected about the line y=x (as they are inverse functions). Therefore the area between the x^1/n, the y axis and the line y=1 must equal the area between x^n, the x-axis and x=1. So this area, plus the area below x^1/n must equal 1. Well, that's how it looks to me, anyway. Hope this is correct. Thank you! Dorothy Last edited: Jan 20, 2007 5. Jan 20, 2007 ### arildno The two part areas do NOT equal each other, but their sum is, indeed 1. 6. Jan 20, 2007 ### Dorothy Weglend But how is that possible, if they are mirror images, the area above x^1/n must equal the area below x^n. Otherwise I don't see how they can sum to 1. What am I missing? 7. Jan 20, 2007 ### arildno Draw the unit square, and the curve y=x^n through it. This partitions the unit square in two areas whose sum are one. The area bounded by the x-axis, y=x^n and x=1 equals 1/(n+1) The area bounded by the y-axis, the curve x=y^(1/n) and y=1 is n/(n+1) Note that x=y^(1/n) denotes exactly the same curve as y=x^n. The reflection about the line y=x occurs when you go from describing the SAME curve equivalently as [itex]y=f(x)$ or $x=f^{-1}(y)$ to describing the different curve $y=f^{-1}(x)$ (equivalently described by x=f(y))

Last edited: Jan 20, 2007
8. Jan 20, 2007

### Dorothy Weglend

I'm not sure why you changed the variables here. It still seems to me the areas must be equal. The original equations, y=x^n and y=x^1/n. the second one is equivalent to y^2 = x.

Therefore the distance from the x axis to the curve y=x^2 must be the same as the distance from the y axis to the curve y^2 = x:

y=(0.5^2) = 0.25, distance from x axis to curve is 0.25
x = (0.5^2) = 0.25, distance from y axis to curve is 0.25

Therefore the areas must be equal (That is, the area above x=y^2 must equal the area below y=x^2).

Or another way to view it: The area beneath the curve x^1/n is n/(n+1). The area above it is 1 - n/(n+1) = 1/(n+1), which is the same as the area below the curve x^n.

This also implies that the curves of x^n and x^(1/n) are reflected about y=x.

In fact, I think we are both saying the same thing.

Dorothy

9. Jan 20, 2007

### arildno

Okay, I misread a bit what you said.
The above statement (1) is correct, statement (2) is incorrect.
In addition to those two areas, you have the area of the region symmetrical about y=x, bounded by the curves y=x^n and y=x^1/n

10. Jan 20, 2007

### Dorothy Weglend

Ah, thanks arildno. And I also mistyped in my first post, I will correct that now.

Thanks again for all your help.