Take the integral and then plug in the upper value and then the lower

[XodoX]

Homework Statement

d/dx _x∫^π sin(t²)

Homework Equations

The Attempt at a Solution

Generally, I take the integral and then plug in the upper value and then the lower value, and then subtract them, correct?

I tried to get help online, but that's a weird result:

http://www.wolframalpha.com/input/?i=ntegral+of+sin%28x^2%29

And with π, it's xsin(π). I don't know how to get the solution here.

 

[Mark44]

Homework Statement

d/dx _x∫^π sin(t²)

Homework Equations

The Attempt at a Solution

Generally, I take the integral and then plug in the upper value and then the lower value, and then subtract them, correct?

I tried to get help online, but that's a weird result:

http://www.wolframalpha.com/input/?i=ntegral+of+sin%28x^2%29

And with π, it's xsin(π). I don't know how to get the solution here.

Here's your integral in LaTeX:
$$d/dx \int_x^{\pi} sin(t^2) dt$$

Use the Fundamental Theorem of Calculus to evaluate this expression. Don't try to find an antiderivative. I assume you're working out of a calculus textbook - there should be some examples of working with this theorem.

 

[DryRun]

Normally, i would first integrate it, evaluate and then differentiate with respect to x. It's quite simple if done that way.

 

[HallsofIvy]

Amazing! You have a simple integral for \int sin(t^2) dt? What is it?

The problem is close to trivial the way Mark44 suggests. If you rewrite it as
-\int_\pi^x sin(t^2) dt
it is trivial using the 'Fundamental Theorem of Calculus'.

 

[DryRun]

Oh, well, i guess I've gone a little bit rusty since the May exams.

My ignorant suggestion would have been: $\frac{-\cos (t^2)}{2t}$

But i guess it's wrong. BTW, there was no need for the intimidating sarcasm. I am aware that you are very knowledgeable.

 

[Curious3141]

Normally, i would first integrate it, evaluate and then differentiate with respect to x. It's quite simple if done that way.

Not in this case. FTC is the way to go.

 

[Curious3141]

Oh, well, i guess I've gone a little bit rusty since the May exams.

My ignorant suggestion would have been: $\frac{-\cos (t^2)}{2t}$

But i guess it's wrong. BTW, there was no need for the intimidating sarcasm.

What works for differentiation (like Chain Rule), is not, in general, immediately reversible for integration.

 

[Curious3141]

Mod note: Removed the quoted text, which gave too much help to the OP.[/color]
The answer is right. But we try not to give out solutions in this forum, and it would've been better for the OP to have worked it out himself.

 

[DryRun]

I know, but i just put it out here to help out the OP over my previous apparently wrong suggestion, and also refreshed my memory after some time away from my maths books. Thanks for pointing out the mistake in my solution presentation. I had it in mind just like you suggested but kept the worked steps to a bare minimum and unintentionally introduced that mistake in the process.

By the way, here is how i had originally pictured the solution:
$$\int^\pi_x \sin (t^2)\,.dt = \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x}$$
$$\frac{d}{dx}\left( \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x} \right)=0-\frac{\sin (x^2).2x}{2x}=-\sin (x^2)$$
It gives the same answer. To be honest, it's quite simple, which is what i was originally referring to.

 

[klondike]

There are 2 parts in FTC. The particular one that's useful to this problem is
f(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt.
Of course, it's trivial that \int_{x}^{a}f(t)dt=-\int_{a}^{x}f(t)dt.

From FTC, F(x) is an anti-derivative of $sin(x^2)$
$$\frac{d}{dx}\left( \int^\pi_x sin(t^2)\,.dt\right)=F'(\pi)-F'(x)=0-F'(x)=-\sin(x^2)$$

 

[Mark44]

I know, but i just put it out here to help out the OP over my previous apparently wrong suggestion, and also refreshed my memory after some time away from my maths books. Thanks for pointing out the mistake in my solution presentation. I had it in mind just like you suggested but kept the worked steps to a bare minimum and unintentionally introduced that mistake in the process.

By the way, here is how i had originally pictured the solution:
$$\int^\pi_x \sin (t^2)\,.dt = \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x}$$
$$\frac{d}{dx}\left( \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x} \right)=0-\frac{\sin (x^2).2x}{2x}=-\sin (x^2)$$
It gives the same answer. To be honest, it's quite simple

But it's wrong, so simplicity is no virtue. The antiderivative of sin(t²) is not -cos(t²)/(2t).

, which is what i was originally referring to.

 

[DryRun]

I realize my mistake now. It is a somewhat special function. To check, i ran $sin (x^2)$ into wolfram integrator and it turns out that it is a Fresnel integral.

 

[klondike]

I realize my mistake now. It is a somewhat special function.

Had you made only one mistake, you won't get the right answer. Kinda like (-1)*(-1)=1.

You have made TWO errors.
1. antiderivative of sin(t²) is wrong as Mark suggested.
2. \frac{d}{dx}\left(\frac{cos(x^2)}{2x}\right)\neq -\sin x^2. Review quotient rule.

I know, but i just put it out here to help out the OP over my previous apparently wrong suggestion, and also refreshed my memory after some time away from my maths books. Thanks for pointing out the mistake in my solution presentation. I had it in mind just like you suggested but kept the worked steps to a bare minimum and unintentionally introduced that mistake in the process.

By the way, here is how i had originally pictured the solution:
$$\int^\pi_x \sin (t^2)\,.dt = \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x}$$
$$\frac{d}{dx}\left( \frac{-\cos (\pi^2)}{2\pi}+\frac{\cos (x^2)}{2x} \right)=0-\frac{\sin (x^2).2x}{2x}=-\sin (x^2)$$
It gives the same answer. To be honest, it's quite simple, which is what i was originally referring to.

 

[XodoX]

It's got to be -cos(x^2)

And

d/dx ₂∫^xsin(t²) dt is also -cos(x²) ?

By the way, I was looking for an explanation. I know it's in the book. Looking at equations shown without explanation didn't help.

 

[Robert1986]

It's got to be -cos(x^2)

And

d/dx ₂∫^xsin(t²) dt is also -cos(x²) ?

By the way, I was looking for an explanation. I know it's in the book. Looking at equations shown without explanation didn't help.

No.

The explanation IS FTC.

Let F'(x) = f(x). Then \int_{a}^{x}f(t)dt = F(x)-F(a). Also, f must be continuous.

Thus, \frac{d}{dx}\int_{a}^{x}f(t)dt = \frac{d}{dx}(F(x)-F(a)).

OK, now try that. Are you familiar with this theorem? If not, you need to be. It is one of the most important things to take out of your calculus class. If you are not, look at the Wikipedia article on the Fundamental Theorem of Calculus. Also, your book should have an explanation.

 

[XodoX]

I already saw that.

 

[Robert1986]

I already saw that.

I'm sorry, I just don't know how it can be made any clearer. Try to identify the parts of the FTC in your question. Actually write it down. Then, if you still don't get it, post what you have done.

 

[Mark44]

I already saw that.

You have been given a strong hint on how to proceed. Don't ask us to explain something like the Fundamental Thm of Calculus, which is in pretty much every calculus textbook.

If you are waiting for us to do the work for you, you will have a long wait. That's not how things work here at PF. We'll steer you in the right direction, but you need to do the work.

 

[XodoX]

We'll steer you in the right direction

Not really.

 

[Mark44]

Not really.

How can you say that? You were going in completely the wrong direction when you attempted to evaluate the integral as the first step. The right direction is getting you to realize that this problem is an application of the Fund. Thm. of Calculus, which has been stated numerous times in this thread.

That should be all the hint you need. The next move is yours. Crack your book open and look at the section that deals with this theorem, and whatever examples it has. If you have specific questions about that, then ask them.

 

[Robert1986]

Not really.

As I said, identify the parts of FTC in your problem. In particular, in your problem, what is F and what is f?

 

[unscientific]

Oh, well, i guess I've gone a little bit rusty since the May exams.

My ignorant suggestion would have been: $\frac{-\cos (t^2)}{2t}$

But i guess it's wrong. BTW, there was no need for the intimidating sarcasm. I am aware that you are very knowledgeable.

I have a suggestion if you don't mind! Whenever I'm unsure whether i did the integral correctly, i will try to differentiate it back to see if it gets back to the original expression..