Composite function and definite integration

[Saitama]

Homework Statement

$$f(x)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}$$
find$$\int_{\frac{1}{4}}^{\frac{3}{4}} f(f(x))dx$$

Homework Equations

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!

 

[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.

 

[Saitama]

Hey ILS!

Is this about a multiple choice perhaps?

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

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 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?

 

[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

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.

 

[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

 

[SammyS]

Homework Statement

$$f(x)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}$$find$$\int_{\frac{1}{4}}^{\frac{3}{4}} f(f(x))dx$$

Homework Equations

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!

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.

 

[SammyS]

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 .

$\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx$

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

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

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

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

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

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

 

[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.

 

[SammyS]

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.

Yes, that's correct.

It is the reciprocal of a whole number.

 

[Saitama]

Woops, sorry for a late reply.

Is there a typo? I do not think that integral is an integer.

Sorry, there is a typo. The reciprocal of the definite integral is the answer.

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

Yes, i am familiar with that. Thanks for that, this might be useful to estimate the answer.

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.

Nicely explained SammyS.

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 .

$\displaystyle \int_{1/4}^{3/4}f(f(x))\,dx$

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

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

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

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

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

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

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.