# A question about the derivation of upper bound integrals

• KungPeng Zhou
KungPeng Zhou
Homework Statement
If f(x)=\int_{0}^{x}x^{2}\sin(t^{2})dt，find \frac{df(x)}{dx}
Relevant Equations
The Substitution Rule for Difinite Integrals，
The Fundamental Theorem of Calculus,
The Product Rule
From the question，we know that the variable is x
First，we can get
f(x)=x^{2}\int_{0}^{x}\sin(t^{2})dt，then\frac{df(x)}{dx}=2x\int_{0}^{x}\sin(t^{2})dt+x^{2}sint^{2}，but I can't deal with \int_{0}^{x}\sin(t^{2})dt，If I do the second differentiation, I can indeed deal with integrals, but there is a second derivative, and the problem is more complicated

Hello, and

##\LaTeX## note: enclose in-line math in double # and displayed math in double $KungPeng Zhou said: Homework Statement: If ##\displaystyle {f(x)=\int_{0}^{x}x^{2}\sin t^{2}\,dt} ##，find ##\frac{df(x)}{dx}## Relevant Equations: The Substitution Rule for Definite Integrals， The Fundamental Theorem of Calculus, The Product Rule From the question，we know that the variable is x First，we can get $$f(x)=x^{2}\int_{0}^{x}\sin t^{2}\,dt，$$then $$\frac{df(x)}{dx}=2x\int_{0}^{x}\sin(t^{2})dt+x^{2}\sin t^{2},$$but I can't deal with $$\int_{0}^{x}\sin(t^{2})dt$$ If I do the second differentiation, I can indeed deal with integrals, but there is a second derivative, and the problem is more complicated Not clear to me what you mean with 'If I do the second differentiation'. The integral $$\int_{0}^{x}\sin(t^{2})dt$$ has a name: Fresnel integral ##\ ## PeroK BvU said: Hello, and ##\LaTeX## note: enclose in-line math in double # and displayed math in double$

Not clear to me what you mean with 'If I do the second differentiation'.
The integral $$\int_{0}^{x}\sin(t^{2})dt$$ has a name: Fresnel integral

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Thanks for the answer, but I'm just a beginner in calculus, do we have an easier way to solve it?

Well, it is the solution; I don't have an easier way to express it ...

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BvU said:
Well, it is the solution; I don't have an easier way to express it ...

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The question is from Calculus by James Stewart(seventh edition).This problem arises in the previous chapters, we have just learned differentiation, we have just reached integration, we have not yet entered multivariate calculus, and we have not learned series.

KungPeng Zhou said:
The question is from Calculus by James Stewart(seventh edition).This problem arises in the previous chapters, we have just learned differentiation, we have just reached integration, we have not yet entered multivariate calculus, and we have not learned series.
It looks quite advanced for a "beginner" problem.

PeroK said:
It looks quite advanced for a "beginner" problem.
In fact，I had a strange question long ago，when I was just starting to learn differential calculus.You can see it in the attach files（Question13)

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PeroK
KungPeng Zhou said:
In fact，I had a strange question long ago，when I was just starting to learn differential calculus.You can see it in the attach files（Question13)
Tough. Looks like you'd need a bit of inspiration for that one!

PeroK said:
Tough. Looks like you'd need a bit of inspiration for that one!
Yes, the first time I saw this question, I had no idea at all， but I spent a day thinking.Suddenly，a brilliant ideal flooded into my head while I was eating.Seeing the answer in the attach files

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• IMG_20230827_191926.jpg
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PeroK
KungPeng Zhou said:
The question is from Calculus by James Stewart(seventh edition).This problem arises in the previous chapters, we have just learned differentiation, we have just reached integration, we have not yet entered multivariate calculus, and we have not learned series.
'Fresnel function' can be found in the index (at least in my 5th edition of Stewart). Haven't found your particular exercise, but from the context I conclude you are not expected to evaluate any further than you did.Re #7 and #9: please post pictures upright -- saves our poor old necks -- and crop irrelevant parts (question 11 and 12).
Question 13 requires some first principles, and you did just fine !

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KungPeng Zhou
BvU said:
'Fresnel function' can be found in the index (at least in my 5th edition of Stewart). Haven't found your particular exercise, but from the context I conclude you are not expected to evaluate any further than you did.Re #7 and #9: please post pictures upright -- saves our poor old necks -- and crop irrelevant parts (question 11 and 12).
Question 13 requires some first principles, and you did just fine !

##\ ##
Oh，I'm sorry.Now I am going to think about using what I have learned to do the previous question.I can't sleep until the problem is solveed

BvU said:
but from the context I conclude you are not expected to evaluate any further than you did.
Shouldn't they substitute ##t=x## for the derivative on the far right?

$$f(x) = x^2 \int_0^x \sin \left(t^2 \right) dt$$

$$\frac{df}{dx}= \frac{df}{dx} \left( x^2 \int_0^x \sin \left(t^2 \right) dt \right)$$

$$= \frac{d x^2}{dx}\int_0^x \sin \left(t^2 \right) dt + x^2 \frac{d}{dt} \left( \int_0^x \sin \left(t^2 \right) dt \right) \frac{dt}{dx}$$

$$= 2x \int_0^x \sin \left(t^2 \right) dt + x^2 \sin \left( t^2 \right)\cdot 1$$

$$= 2x \int_0^x \sin \left(t^2 \right) dt + x^2 \sin \left( x^2 \right)$$

Or do I have some lose screw?

erobz said:
Or do I have some lose screw?
That would be "loose". I wouldn't go so far as to say you have a loose screw, but your notation is incorrect and you have an error in your work.
erobz said:
$$f(x) = x^2 \int_0^x \sin \left(t^2 \right) dt$$

$$\frac{df}{dx}= \frac{df}{dx} \left( x^2 \int_0^x \sin \left(t^2 \right) dt \right)$$
The right side of the equation should start with ##\frac d{dx}##, not ##\frac{df}{dx}##. I.e., it should be the differentiation operator, not the derivative of f.
You're differentiating the integral in parentheses, not performing a multiplication.
erobz said:
$$= \frac{d x^2}{dx}\int_0^x \sin \left(t^2 \right) dt + x^2 \frac{d}{dt} \left( \int_0^x \sin \left(t^2 \right) dt \right) \frac{dt}{dx}$$
The second term on the right side above is incorrect. It should be
$$x^2 \frac{d}{dx} \left( \int_0^x \sin \left(t^2 \right) dt \right)$$
This simplifies, using the Fundamental Theorem of Calculus, to
##x^2 \sin(x^2)##.

Your final result looks correct to me, but the intervening work is off.

Mark44 said:
That would be "loose". I wouldn't go so far as to say you have a loose screw, but your notation is incorrect and you have an error in your work.
The right side of the equation should start with ##\frac d{dx}##, not ##\frac{df}{dx}##. I.e., it should be the differentiation operator, not the derivative of f.

Yes, I just didn't notice I put ##f## in there.
Mark44 said:
You're differentiating the integral in parentheses, not performing a multiplication.
The second term on the right side above is incorrect. It should be
$$x^2 \frac{d}{dx} \left( \int_0^x \sin \left(t^2 \right) dt \right)$$
This simplifies, using the Fundamental Theorem of Calculus, to
##x^2 \sin(x^2)##.

So then we are not really using the Chain Rule like I have in that way?

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erobz said:
So then we are not really using the Chain Rule like I have in that way?
Correct. One part of the FTC says that ##\frac d{dx} \int_a^x f(r)dr = f(x)##. That's what I used -- no chain rule at all.

Mark44 said:
Correct. One part of the FTC says that ##\frac d{dx} \int_a^x f(r)dr = f(x)##. That's what I used -- no chain rule at all.
I was thinking if we let:

$$G(t) = \int \sin \left( t^2 \right) dt$$

Then we would have:

$$\frac{dG}{dx} = \frac{dG(x)}{dt} \frac{dt}{dx} - \cancel{\frac{dG(a)}{dt} \frac{dt}{da}}^0$$

The last term being ##0## because ##\frac{dt}{da} = 0 ##.

Is that nonsense? What do we do if the lower limit is also a function of ##x##, not just the upper limit?

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erobz said:
I was thinking if we let:

$$G(t) = \int \sin \left( t^2 \right) dt$$

Then we would have:

$$\frac{dG}{dx} = \frac{dG(x)}{dt} \frac{dt}{dx} - \cancel{\frac{dG(a)}{dt} \frac{dt}{da}}^0$$

The last term being ##0## because ##\frac{dt}{da} = 0 ##.

Is that nonsense?
I think so, or at the least, not helpful.
The left side of your equation is ##\frac{dG}{dx}##. That would be ##\frac{dG(t)}{dx}##. I don't see it as being valid to switch immediately to ##\frac{dG(x)}{dt}## as you have on the right side. Also, you're flipping between indefinite and definite integrals (or at least, not showing the limits of integration, which makes it harder to follow).
Either way, use the FTC as I showed earlier.
erobz said:
What do we do if the lower limit is also a function of ##x##, not just the upper limit?
Split the integral into two integrals, with each having one of the functions of x in one or the other integration limits. This is standard stuff in any calculus textbook where they discuss the FTC.

WWGD and PeroK
erobz said:
I was thinking if we let:

$$G(t) = \int \sin \left( t^2 \right) dt$$

Then we would have:

$$\frac{dG}{dx} = \frac{dG(x)}{dt} \frac{dt}{dx} - \cancel{\frac{dG(a)}{dt} \frac{dt}{da}}^0$$

The last term being ##0## because ##\frac{dt}{da} = 0 ##.

Is that nonsense? What do we do if the lower limit is also a function of ##x##, not just the upper limit?
Since you define:$$G\left(t\right)\equiv\intop\sin\left(t^{2}\right)dt$$it follows by the fundamental theorem of integral calculus that:$$G\left(b\right)-G\left(a\right)=\intop_{a}^{b}\sin\left(t^{2}\right)dt$$This is true even if ##a,b## are functions of some new variable ##x##, so the answer to your last question is:$$\intop_{a\left(x\right)}^{b\left(x\right)}\sin\left(t^{2}\right)dt=G\left(b\left(x\right)\right)-G\left(a\left(x\right)\right)$$

PeroK
renormalize said:
Since you define:$$G\left(t\right)\equiv\intop\sin\left(t^{2}\right)dt$$
It's slightly more precise to let ##G## be a function such that$$G'(t) = \sin(t^2)$$This avoids the technicality that an indefinite integral is an equivalent class of functions. Then
renormalize said:
It follows by the fundamental theorem of integral calculus that:$$G\left(b\right)-G\left(a\right)=\intop_{a}^{b}\sin\left(t^{2}\right)dt$$This is true even if ##a,b## are functions of some new variable ##x##, so the answer to your last question is:$$\intop_{a\left(x\right)}^{b\left(x\right)}\sin\left(t^{2}\right)dt=G\left(b\left(x\right)\right)-G\left(a\left(x\right)\right)$$

renormalize
Mark44 said:
I think so, or at the least, not helpful.
The left side of your equation is ##\frac{dG}{dx}##. That would be ##\frac{dG(t)}{dx}##. I don't see it as being valid to switch immediately to ##\frac{dG(x)}{dt}## as you have on the right side. Also, you're flipping between indefinite and definite integrals (or at least, not showing the limits of integration, which makes it harder to follow).
Either way, use the FTC as I showed earlier.
Split the integral into two integrals, with each having one of the functions of x in one or the other integration limits. This is standard stuff in any calculus textbook where they discuss the FTC.
Perhaps my notation is so abusive it is not interpretable by the pros!

Let me tweak the integral so its explicitly solvable and try to be more careful about my notation:

Let

$$G(t) = \int_{t_l = 2x}^{t_u=x^2} \sin ( t ) dt$$

It follows that:

$$G(t) = - \left. \cos(t) \right|_{t_l=2x}^{t_u=x^2}$$

$$G(x) = \cos ( 2x ) - \cos ( x^2 )$$

Then

\begin{aligned} \frac{dG}{dx} &= \frac{dG(t_u)}{dt_u} \frac{dt_u}{dx} - \frac{dG(t_l)}{dt_l} \frac{dt_l}{dx} \\ \quad \\ &= 2x \sin ( x^2 ) - 2 \sin ( 2x ) \end{aligned}

Any better?

If not I'll give up trying to salvage it.

erobz said:
Perhaps my notation is so abusive it is not interpretable by the pros!

Let me tweak the integral so its explicitly solvable and try to be more careful about my notation:

Let

$$G(t) = \int_{t_l = 2x}^{t_u=x^2} \sin ( t ) dt$$
This is not valid. If ##t## is your dummy integration variable, it cannot appear outside the integrand. That ##G## is a function of ##x## and ##x## is not a function of ##t##.

Mark44 and erobz
PeroK said:
This is not valid. If ##t## is your dummy integration variable, it cannot appear outside the integrand. That ##G## is a function of ##x## and ##x## is not a function of ##t##.

$$G = \int_{t_l = 2x}^{t_u=x^2} \sin ( t ) dt$$

Everything else is ok though?

erobz said:

$$G = \int_{t_l = 2x}^{t_u=x^2} \sin ( t ) dt$$

Everything else is ok though?
Simplest and best is$$G(x) = \int_{2x}^{x^2} \sin ( t ) dt$$Note, however, that is a different function from the ##G## defined above.

erobz
erobz said:
Let Let
$$G(t) = \int_{t_l = 2x}^{t_u=x^2} \sin ( t ) dt$$
No. As already mentioned by @PeroK, G is a function of x, not t.

Let ##G(x) = \int_{2x}^{x^2} \sin ( t ) dt##
##= \int_{2x}^a \sin ( t ) dt + \int_a^{x^2} \sin ( t ) dt##
##= -\int_a^{2x} \sin ( t ) dt+ \int_a^{x^2} \sin ( t ) dt##
##= \left.\cos(t)\right |_a^{2x} - \left.\cos(t)\right |_a^{x^2}##
##=\cos(2x) - \cos(a) - \cos(x^2) + \cos(a) = \cos(2x) - \cos(x^2)##

erobz

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