# Composite function and definite integration

1. Nov 3, 2012

### Saitama

1. The problem statement, all variables and given/known data
$$f(x)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}$$
find$$\int_{\frac{1}{4}}^{\frac{3}{4}} f(f(x))dx$$

2. Relevant equations

3. The attempt at a solution
I am clueless here. I started by writing f(f(x)) as
$$(f(x))^3-\frac{3(f(x))^2}{2}+f(x)+\frac{1}{4}$$
I don't think expanding (f(x))^3 would be of any use because this question is from a test paper and it would take a long time if i expanded (f(x))^3 and the other terms. I don't have any idea how to proceed. I observed that the f(x) is always an increasing function, therefore only one real root for f(x) exist but i don't think that would be of any use here.

Any help is appreciated!

2. Nov 3, 2012

### I like Serena

Hey Pranav-Arora!

Is this about a multiple choice perhaps?

Then you'd be supposed to make an estimate (using inscribed rectangles) and match the answer.
Or alternatively make a drawing and measure the area.

I agree that expanding everything is rather time consuming.

3. Nov 3, 2012

### Saitama

Hey ILS!

Nope, this question was from a section where the answer is an integer from 0 to 9.

I don't understand why are you talking about rectangles here. Maybe i could estimate the area using the graph separately for (f(x))^3 and (f(x))^2.

Isn't there any proper method to get an exact answer?

4. Nov 4, 2012

### I like Serena

Ah, so you do have at least some limitation on the acceptable answers.

An integral is the area under a graph.

A first estimate is the area of the rectangle that has a height equal to a point on the graph, and a width equal to the boundaries.
Somewhat like this (for an integral with boundaries 0.5 and 1.5):
http://www.livemath.com/hoffnercalculus/preview/Chapter%206/Chpt6_Resources/image37.gif [Broken]

In your case you could take f(f(1/2)) as the height and (3/4 - 1/4) as the width.
I believe that will suffice to find the proper answer, since that has to be an integer.

The method to get the exact answer is exactly what you were already doing.
I do not see ways to significantly speed up that process.

Last edited by a moderator: May 6, 2017
5. Nov 4, 2012

### lurflurf

Is there a typo? I do not think that integral is an integer.
Are you familiar with

$$(b-a)L\le \int_a^b f(x) dx \le (b-a)U$$
where U and L are numbers such that
L<f<U

6. Nov 4, 2012

### SammyS

Staff Emeritus
There are several things to notice here.

The point, (1/2, 1/2), is a fixed point. So, f(1/2) = 1/2, and f(f(1/2)) = 1/2 .

Also, the point, (1/2, 1/2) is a point of inflection for f(x), and f(x) is a cubic function, so the graph, y = f(x) is invariant under a 180° rotation about the fixed point, (1/2, 1/2). In other words, if you shift the graph, y = f(x), to the left by 1/2 and down by 1/2, the resulting graph, y = f(x + 1/2) - 1/2 is symmetric with respect to the origin.

Thus, the function, g(x) = f(x + 1/2) - 1/2 is an odd function, actually an odd cubic function.

The function, g(g(x)), is also an odd function, so $\displaystyle \int_{-1/4}^{1/4}g(g(x))\,dx=0\ .$

You can also show, graphically or otherwise, that y=g(g(x)) is a shifted version of y=f(f(x)) .

So I like Serena's idea of using a rectangle is relevant.

7. Nov 5, 2012

### SammyS

Staff Emeritus
Better yet:

Define $\displaystyle g(x) = f(x+(1/2)) - 1/2$, which is an odd (cubic) function.

Now for the integral $\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx\ :$

Do the substitution x = u + 1/2, then dx = du. x = 1/4, 3/4 becomes u = -1/4, 1/4 .

$\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx$
$\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))\,du$

$\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))-\frac{1}{2}+\frac{1}{2}\,du$

$\displaystyle =\int_{-1/4}^{1/4}f(f(u + 1/2))-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du$

$\displaystyle =\int_{-1/4}^{1/4}f\left(f(u + 1/2)-1/2 +\frac{1}{2}\right)-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du$

$\displaystyle =\int_{-1/4}^{1/4}f\left(g(u) +\frac{1}{2}\right)-\frac{1}{2}\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du$

$\displaystyle =\int_{-1/4}^{1/4}g(g(u))\,du+\int_{-1/4}^{1/4}\frac{1}{2}\,du$

8. Nov 5, 2012

### I like Serena

I did not analyze as far as you did, but if I understand you correctly, the area of the rectangle is equal to the exact solution. ;)
And indeed, that is not a whole number.

9. Nov 5, 2012

### SammyS

Staff Emeritus
Yes, that's correct.

It is the reciprocal of a whole number.

10. Nov 5, 2012

### Saitama

Woops, sorry for a late reply.

Sorry, there is a typo. The reciprocal of the definite integral is the answer.
Yes, i am familiar with that. Thanks for that, this might be useful to estimate the answer.

Nicely explained SammyS.

Excellent! How do they expect us to think like this and that too in the examination room.
The solution would be take me some time to understand it completely, i will get back to this thread if i get stuck.

I will mention it again, there's a typo. The answer is the reciprocal of the value of the definite integral.

Last edited: Nov 5, 2012