Solve Integral 'Riddles': Show ∫g(x) dx ≠ ∫g(x^2) 2xdx

[Justabeginner]

Homework Statement

Show that ∫g(x) dx = ∫g(x^2) 2xdx with a=0 and b=1

The followup question asks me to prove that the above integrals are NOT equal, by providing two functions that disprove it (one where the former integral is larger, and one where the latter integral is larger).

The last question with this set asks which of the previous two integrals is larger if g(x) is increasing on (0,1)?

Homework Equations

The Attempt at a Solution

I used g(x)= x and I got u= x^2
du= 2x dx

1 | x^2/2
0 |

1 | u^2/2
0 | x^4/2

1/2 = 1/2

So for the former integral being larger, my example is g(x)= 3x + 6. For the latter integral being larger, my example is g(x)= -(x^2) + 4.

For the last question in the set, I have no clue what to do, because only 3x+ 6 is increasing on that interval. In fact, -(x^2) + 4 is decreasing, so I'm stumped. Help please? Thank you.

 

[Mentz114]

The question seems incomplete - where do a and b appear ? are they the limits of integration ?

 

[Justabeginner]

Yes they are the limits of integration. Sorry about that!

 

[LCKurtz]

Homework Statement

Show that ∫g(x) dx = ∫g(x^2) 2xdx with a=0 and b=1

Do you mean $\int_0^1g(x)\, dx = \int_0^1 g(x^2)2x\, dx$?

The followup question asks me to prove that the above integrals are NOT equal, by providing two functions that disprove it (one where the former integral is larger, and one where the latter integral is larger).

So you are asked to prove it is true and then also asked to give counterexamples to its truth? I suggest you need to quote the exact and complete statement of the problem.

 

[Justabeginner]

Yes that's what the second question asks.

Show that the integral of g(x) dx is not equal to the integral of g(x^2) dx when a =0 and b=1 by finding two examples, one where the former integral is larger, and the other where the latter integral is larger. Give explicit formulae for your functions.

 

[Mentz114]

Yes they are the limits of integration. Sorry about that!

No problem. This a bit better
∫_a^bg(x) dx = ∫_a^bg(x^2) 2x dx

I still don't get how you can be asked to prove the equality, and then to *disprove* it.

 

[Justabeginner]

Well they are two parts to the same question, so I guess that's what it is.

 

[LCKurtz]

Are you sure you aren't supposed to show it is true when a = 0 and b = 1 but false for other values of a and b? Like I asked earlier, please give a word-for-word statement of the problem. Otherwise you are just wasting our time.

 

[Justabeginner]

Yes that's what the second question asks.

Show that the integral of g(x) dx is not equal to the integral of g(x^2) dx when a =0 and b=1 by finding two examples, one where the former integral is larger, and the other where the latter integral is larger. Give explicit formulae for your functions.

This is the EXACT wording of the question. Sorry if I'm wasting your time...

And no it is not asking for me to change the upper and lower limits of the integral, but it's asking me to select different functions altogether.

 

[Mandelbroth]

Homework Statement

Show that ∫g(x) dx = ∫g(x^2) 2xdx with a=0 and b=1

The followup question asks me to prove that the above integrals are NOT equal, by providing two functions that disprove it (one where the former integral is larger, and one where the latter integral is larger).

The last question with this set asks which of the previous two integrals is larger if g(x) is increasing on (0,1)?

$$\int_0^1 g(x) \, dx = \int_0^1 g(x^2) 2x \, dx = \int_0^1 g(u) \frac{du}{dx} \, dx = \int_0^1 g(u) \, du$$

What went wrong with that?

 

[slider142]

Yes that's what the second question asks.

Show that the integral of g(x) dx is not equal to the integral of g(x^2) dx when a =0 and b=1 by finding two examples, one where the former integral is larger, and the other where the latter integral is larger. Give explicit formulae for your functions.

The bolded part is very important. It is not the same integral as in the original question.

 

[Justabeginner]

Yes slider142. That's exactly what made me rethink if I had even done this correctly. Thank you for bolding that.

 

[HallsofIvy]

slider142's point is that first problem asks you to show that
$\int_0^1 g(x)dx= \int_0^1 g(x^2) 2x dx$
while the second asks you to give examples to show that $\int_0^1 g(x)dx$ is NOT the same as $\int_0^1 g(x^2)dx$.

Do you see the difference? There is no "2x" in the second.

 

[LCKurtz]

And that's why I asked for the exact statement of the problem. You didn't have it stated correctly.

 

[Justabeginner]

I realized that the problem is stated that way LCKurtz, and that everyone wondered why it was so, which is why I retyped the whole thing out to clarify that that is exactly how it was worded. I solved it despite that, and I'm almost 1000% certain it's wrong, which is why I decided to ask the experts on here.

 

[haruspex]

I realized that the problem is stated that way LCKurtz, and that everyone wondered why it was so, which is why I retyped the whole thing out to clarify that that is exactly how it was worded. I solved it despite that, and I'm almost 1000% certain it's wrong, which is why I decided to ask the experts on here.

Not sure I understand what you're saying there. With the exact wording, the question makes sense. What is it that you are almost a thousand percent sure is wrong?

 

[LCKurtz]

I realized that the problem is stated that way LCKurtz, and that everyone wondered why it was so, which is why I retyped the whole thing out to clarify that that is exactly how it was worded. I solved it despite that, and I'm almost 1000% certain it's wrong, which is why I decided to ask the experts on here.

That the problem was stated what way?

This confuses me too. I can't tell if you understand yet that your original post did not state the problem correctly.

 

[Justabeginner]

I was a 1000% percent sure that MY solution to the problem is incorrect. The problem was stated in the way slider142 presented it: that the integral in the second part of the problem is NOT the same as the integral in the first part of the problem. This is where I'm stuck. That's why I wanted to ask all of you experts, if my solution is right or not.

 

[LCKurtz]

Homework Statement

Show that ∫g(x) dx = ∫g(x^2) 2xdx with a=0 and b=1

The followup question asks me to prove that the above integrals are NOT equal, by providing two functions that disprove it (one where the former integral is larger, and one where the latter integral is larger).

The last question with this set asks which of the previous two integrals is larger if g(x) is increasing on (0,1)?

Homework Equations

The Attempt at a Solution

I used g(x)= x and I got u= x^2
du= 2x dx

1 | x^2/2
0 |

1 | u^2/2
0 | x^4/2

1/2 = 1/2

So for the former integral being larger, my example is g(x)= 3x + 6. For the latter integral being larger, my example is g(x)= -(x^2) + 4.

For the last question in the set, I have no clue what to do, because only 3x+ 6 is increasing on that interval. In fact, -(x^2) + 4 is decreasing, so I'm stumped. Help please? Thank you.

I was a 1000% percent sure that MY solution to the problem is incorrect. The problem was stated in the way slider142 presented it: that the integral in the second part of the problem is NOT the same as the integral in the first part of the problem. This is where I'm stuck. That's why I wanted to ask all of you experts, if my solution is right or not.

Looking at your solution above, for part a you have shown that it works for $g(x) = x$. But you were to show it for all $g(x)$. So no, your solution for that one isn't correct.

For your solutions to one or the other being larger, instead of just giving answers, show us what you did. I don't know whether they are right or not without working them myself. You shouldn't expect us to work them out for ourselves to see if we agree. Just show us what you did.

 

[Justabeginner]

Okay, so I would have to choose another function for part a, and show the work for that.

For part B,

g(x) = 3x + 6
integral g(x) = 3/2x^2 + 6x (dx)
integral g(1)= 7.5

integral g(x^2)= x^3 + 6x
integral g(1^2)= 7

Other function:

g(x) = -(x^2) + 4
integral of g(x)= =1/3x^3 + 4x
g(1)= 11/3

integral of g(x^2)= -1/5x^5 + 4x
g(1^2)= 19/5

As for part C, I'm clueless. Thank you.

 

[Mandelbroth]

Okay, so I would have to choose another function for part a, and show the work for that.

Not just any function. For any arbitrary function g(x). That's how you show that it works for all functions g(x).

As for part c, which one is larger if g is increasing on (0,1)?

How does the value of x^2 compare to the value of x if x is between 0 and 1?

 

[Mark44]

Okay, so I would have to choose another function for part a, and show the work for that.

For part B,

g(x) = 3x + 6
integral g(x) = 3/2x^2 + 6x (dx)

It would be very helpful if you learned how to use the mathematics symbols that are available on this site. Under the text entry panel there is a button that says Go Advanced. Click it to see an expanded menu above the text entry pane, as well as a Quick Symbols pane to the right.

The equations above would be correctly written as:
g(x) = 3x + 6
∫g(x)dx = ∫(3x + 6)dx = (3/2)x² + 6x + C (dx should not appear in the antiderivative.)

integral g(1)= 7.5

I'm not sure what you mean by this (i.e. "integral g(1)").
If you define G(x) = (3/2)x² + 6x, then G(1) = 3/2 + 6 = 7.5

integral g(x^2)= x^3 + 6x

I don't know how you got this (above).

integral g(1^2)= 7

Ignoring momentarily that "integral g(x)" is terrible notation, why wouldn't integral g(1^2) be the same as integral g(1)?

Other function:

g(x) = -(x^2) + 4
integral of g(x)= =1/3x^3 + 4x

If you mean ∫g(x)dx, then you have a sign error. Also, you are missing the constant of integration.

g(1)= 11/3

integral of g(x^2)= -1/5x^5 + 4x

?
How did you get this?

g(1^2)= 19/5

As for part C, I'm clueless. Thank you.

What is part C? Nothing you've posted has a part C in it, as far as I can tell.

 

[Justabeginner]

Not just any function. For any arbitrary function g(x). That's how you show that it works for all functions g(x).

As for part c, which one is larger if g is increasing on (0,1)?

How does the value of x^2 compare to the value of x if x is between 0 and 1?

Oh wow! Of course Thank you. X^2 will be less than x on this interval!

 

[Justabeginner]

It would be very helpful if you learned how to use the mathematics symbols that are available on this site. Under the text entry panel there is a button that says Go Advanced. Click it to see an expanded menu above the text entry pane, as well as a Quick Symbols pane to the right.

The equations above would be correctly written as:
g(x) = 3x + 6
∫g(x)dx = ∫(3x + 6)dx = (3/2)x² + 6x + C (dx should not appear in the antiderivative.)

I'm not sure what you mean by this (i.e. "integral g(1)").
If you define G(x) = (3/2)x² + 6x, then G(1) = 3/2 + 6 = 7.5
I don't know how you got this (above).
Ignoring momentarily that "integral g(x)" is terrible notation, why wouldn't integral g(1^2) be the same as integral g(1)?
If you mean ∫g(x)dx, then you have a sign error. Also, you are missing the constant of integration.
?
How did you get this?What is part C? Nothing you've posted has a part C in it, as far as I can tell.

The part C was the question regarding the intervals, but mandelbroth helped me with that part. For 'integral g(1)' I meant the value of the integral of g(x) at the x-value of 1. The value of the integral at g(1^2) would not be the same as the value of the integral at g(1) as the two integrals are entirely different. One has an x^2 and the other is simply an x. I definitely should have been clearer with the way I wrote the problem, and I will make sure to use the symbols available to me. I solved the problem and am good for now. Thank you for all your help!