Determining the area enclosed by inverse

[Saitama]

Homework Statement

Area enclosed by y=g(x), x=-3, x=5 and x-axis where g(x) is inverse function of $f(x)=x^3+3x+1$ is A, then find [A] where [.] denotes the greatest integer function.

Homework Equations

The Attempt at a Solution

Honestly, I see no way to proceed here. Finding out the inverse is not possible here so I guess there is some trick to the problem.

Any help is appreciated. Thanks!

 

[pasmith]

Homework Statement

Area enclosed by y=g(x), x=-3, x=5 and x-axis where g(x) is inverse function of $f(x)=x^3+3x+1$ is A, then find [A] where [.] denotes the greatest integer function.

Homework Equations

The Attempt at a Solution

Honestly, I see no way to proceed here. Finding out the inverse is not possible here so I guess there is some trick to the problem.

Any help is appreciated. Thanks!

By inspection, $f(1) = 5$ and $f(-1) = -3$. Hence $g(-3) = -1$ and $g(5) = 1$. I believe that $A$ is then
$$\int_0^1 5 - f(y)\,dy$$

 

[Saitama]

$$A = \int_{-3}^5 g(x)\,dx = \int_{g(-3)}^{g(5)} 5 - f(y)\,dy$$

How do you get this?

 

[pasmith]

How do you get this?

Ignore that; it's wrong. I edited my post.

 

[Saitama]

Ignore that; it's wrong. I edited my post.

I meant how you get 5-f(y)?

 

[pasmith]

I meant how you get 5-f(y)?

Draw a graph of $y = g(x)$ (which is by definition the curve $x = f(y)$) for $-3 \leq x \leq 5$, and identify the area A. Then express A as an integral with respect to y.

 

[Saitama]

Draw a graph of $y = g(x)$ (which is by definition the curve $x = f(y)$) for $-3 \leq x \leq 5$, and identify the area A. Then express A as an integral with respect to y.

I honestly don't see it. Can you please elaborate some more or give me a relevant link?

 

[I like Serena]

Hi Pranav!

The area enclosed by $f^{-1}(x)$ between $x=-3$ and $x=5$, is the same area as the one enclosed by $f(y)$ between $y=-3$ and $y=5$.
It's what you get if you take the mirror image in the line y=x.

Can you find that area?

 

[Saitama]

Hi ILS! :)

Hi Pranav!

The area enclosed by $f^{-1}(x)$ between $x=-3$ and $x=5$, is the same area as the one enclosed by $f(y)$ between $y=-3$ and $y=5$.
It's what you get if you take the mirror image in the line y=x.

Can you find that area?

$f(y)=y^3+3y+1$.

The area enclosed by f(y) between y=-3 and y=5 is $\displaystyle \int_{-3}^5 (y^3+3y+1)dy=168$.

But this doesn't look like the right answer.

 

[I like Serena]

Hi ILS! :)

$f(y)=y^3+3y+1$.

The area enclosed by f(y) between y=-3 and y=5 is $\displaystyle \int_{-3}^5 (y^3+3y+1)dy=168$.

But this doesn't look like the right answer.

My bad. I meant the area enclosed by f(x) between y=-3 and y=5 (and the y-axis).

 

[Saitama]

My bad. I meant the area enclosed by f(x) between y=-3 and y=5 (and the y-axis).

But for that I need the inverse of f(x).

 

[I like Serena]

But for that I need the inverse of f(x).

Not really. You only need the boundaries for x that correspond to y=-3 and y=5.
So you need to solve f(x)=-3 and f(x)=5.
Can you?

 

[Saitama]

Not really. You only need the boundaries for x that correspond to y=-3 and y=5.
So you need to solve f(x)=-3 and f(x)=5.
Can you?

pasmith mentioned them above. :P

For f(x)=-3, x=-1 and for f(x)=5,x=1.

 

[I like Serena]

pasmith mentioned them above. :P

For f(x)=-3, x=-1 and for f(x)=5,x=1.

So...

 

[Saitama]

So...

Do I have to integrate f(x) from x=-1 to 1?

 

[I like Serena]

Do I have to integrate f(x) from x=-1 to 1?

Not exactly.
Take a look at this Wolfram picture.

You need the area enclosed between the lines x=0, y=-3, and the graph.
And that combined with the area enclosed between the lines x=0, y=5, and the graph.

 

[Saitama]

Not exactly.
Take a look at this Wolfram picture.

You need the area enclosed between the lines x=0, y=-3, and the graph.
And that combined with the area enclosed between the lines x=0, y=5, and the graph.

I am not sure if I get it but I guess I will have to break the integral. I just saw the plot of $x^3+3x+1$ and it looks like there is a root between -1 and 0. Unfortunately, the root is fractional so I don't know would I have noticed that during the examination. I think I am missing something else too.

 

[I like Serena]

I am not sure if I get it but I guess I will have to break the integral. I just saw the plot of $x^3+3x+1$ and it looks like there is a root between -1 and 0. Unfortunately, the root is fractional so I don't know would I have noticed that during the examination. I think I am missing something else too.

Perhaps you can calculate the integral of f(x) between x=-1 and x=1?

Which area will you have calculated?

 

[I like Serena]

For reference, here is the Wolfram picture showing what you have to calculate according to your problem description.

Can you indicate which areas are the relevant areas?

 

[Saitama]

Perhaps you can calculate the integral of f(x) between x=-1 and x=1?

I will break this integral from x=-1 to a and from x=a to 1, where a is the root. The value is negative for -1 to a is negative. Changing the sign and adding it to value obtained between a to 1 gives the area between f(x) and the x-axis. But that's not what I need.

To calculate what is required, I need the value of a but as I said before, it is fractional and would be impossible to find during the exam.

 

[LCKurtz]

Do I have to integrate f(x) from x=-1 to 1?

No. You can do it for x = -1 to 1 if you have the correct integrand(s). Have you drawn a picture of the required area? Show us your integral.

[Edit] Didn't see the second page before I posted this...

 

[I like Serena]

I will break this integral from x=-1 to a and from x=a to 1, where a is the root. The value is negative for -1 to a is negative. Changing the sign and adding it to value obtained between a to 1 gives the area between f(x) and the x-axis. But that's not what I need.

To calculate what is required, I need the value of a but as I said before, it is fractional and would be impossible to find during the exam.

You do not need that "a".
Take another look at the graph: http://m.wolframalpha.com/input/?i=plot[y=x^3+3x+1,x=-1,x=0,x=1,y=-3,y=5,+{x,-1,1}]&x=-1505&y=-71

What you name "a", is the y-value for x=0.
You don't need to integrate to and from "a", but to and from x=0.

 

[Saitama]

You do not need that "a".
Take another look at the graph: http://m.wolframalpha.com/input/?i=plot[y=x^3+3x+1,x=-1,x=0,x=1,y=-3,y=5,+{x,-1,1}]&x=-1505&y=-71

What you name "a", is the y-value for x=0.
You don't need to integrate to and from "a", but to and from x=0.

I have shown the required area in the attachment, the shaded area is what we require, right?

I am not sure but is the graph symmetrical? It looks so.

I calculated the area on the right side of x=0. It can be found from the integral:
$$5-\int_{0}^1 (x^3+3x+1)dx=5-\frac{11}{4}=\frac{9}{4}$$

Twice of 9/4 is 9/2, hence area is 9/2=4.5, or [A]=4.

This is the answer shown by the key. Is this correct?

 

[I like Serena]

I have shown the required area in the attachment, the shaded area is what we require, right?

I am not sure but is the graph symmetrical? It looks so.

I calculated the area on the right side of x=0. It can be found from the integral:
$$5-\int_{0}^1 (x^3+3x+1)dx=5-\frac{11}{4}=\frac{9}{4}$$

Twice of 9/4 is 9/2, hence area is 9/2=4.5, or [A]=4.

This is the answer shown by the key. Is this correct?

Good point.
You can calculate the area to the right of the y-axis, which is what you did.
The graph is not symmetrical, but it's close enough for our purposes.
You are only asked to give an integer approximation, which is what you did.

Good!

 

[LCKurtz]

Good point.
You can calculate the area to the right of the y-axis, which is what you did.
The graph is not symmetrical, but it's close enough for our purposes.
You are only asked to give an integer approximation, which is what you did.

Good!

Well, I would disagree with that. If you don't prove it is symmetric, don't use it. It would be easy enough to calculate the areas of both regions with dx integrals and be sure.

 

[Saitama]

Good point.
You can calculate the area to the right of the y-axis, which is what you did.
The graph is not symmetrical, but it's close enough for our purposes.
You are only asked to give an integer approximation, which is what you did.

Good!

Thanks a lot ILS!

Now that the problem is solved, I was wondering if you could help me understand the solution given in the solution booklet? The solution booklet solves the problem in a single line and I am clueless about what the paper setters did. Here's what is written in booklet:

$$A=\left(4-\int_0^1(f(y)-1)dy \right)+\left|4-\int_{-1}^0 (1-f(y))dy\right|=\frac{9}{2}$$

Any idea what is going on above?

 

[LCKurtz]

Thanks a lot ILS!

Now that the problem is solved, I was wondering if you could help me understand the solution given in the solution booklet? The solution booklet solves the problem in a single line and I am clueless about what the paper setters did. Here's what is written in booklet:

$$A=\left(4-\int_0^1(f(y)-1)dy \right)+\left|4-\int_{-1}^0 (1-f(y))dy\right|=\frac{9}{2}$$

Any idea what is going on above?

That is an unnecessarily obscure way to set it up. The natural way would be just use the usual calculus formula for area between two curves:$$
A = \int_a^b y_{upper} - y_{lower}~dx$$on both pieces:$$
\int_{-1}^0 f(x) -(-3)~dx + \int_0^15-f(x)~dx$$

 

[Saitama]

That is an unnecessarily obscure way to set it up. The natural way would be just use the usual calculus formula for area between two curves:$$
A = \int_a^b y_{upper} - y_{lower}~dx$$on both pieces:$$
\int_{-1}^0 f(x) -(-3)~dx + \int_0^15-f(x)~dx$$

Yep, agreed! :)

But it looks to me that the solution was trying to define a odd function g(x)=f(x)-1. Anyways, its quite difficult to catch that during the exam so simply setting up the integrals is way better.

Do you think it would be possible to solve the problem without looking at the plots? I couldn't get a feel for the problem until I looked at the graphs. Do you have any suggestions?

Thank you very much LCKurtz and ILS! I got to learn something new.

 

[LCKurtz]

Yep, agreed! :)

Do you think it would be possible to solve the problem without looking at the plots? I couldn't get a feel for the problem until I looked at the graphs. Do you have any suggestions?

My advice would be to NEVER set up an area integral without making a plot. The only exception would be where it is so simple that you already have a mental image, which amounts to the same thing.

 

[Saitama]

My advice would be to NEVER set up an area integral without making a plot. The only exception would be where it is so simple that you already have a mental image, which amounts to the same thing.

Thank you once again LCKurtz! :)