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

1. Jul 22, 2012

XodoX

1. The problem statement, all variables and given/known data

d/dx xπ sin(t2)

2. Relevant equations

3. 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.

2. Jul 22, 2012

Staff: Mentor

Re: Integrals

$$d/dx \int_x^{\pi} sin(t^2) dt$$

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.

3. Jul 23, 2012

sharks

Re: Integrals

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

4. Jul 23, 2012

HallsofIvy

Staff Emeritus
Re: Integrals

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'.

5. Jul 23, 2012

sharks

Re: Integrals

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.

6. Jul 23, 2012

Curious3141

Re: Integrals

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

7. Jul 23, 2012

Curious3141

Re: Integrals

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

8. Jul 23, 2012

Curious3141

Re: Integrals

Mod note: Removed the quoted text, which gave too much help to the OP.
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.

Last edited by a moderator: Jul 23, 2012
9. Jul 23, 2012

sharks

Re: Integrals

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.

Last edited: Jul 23, 2012
10. Jul 23, 2012

klondike

Re: Integrals

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 $\int_{x}^{a}f(t)dt=-\int_{a}^{x}f(t)dt$.

11. Jul 23, 2012

Staff: Mentor

Re: Integrals

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

12. Jul 23, 2012

sharks

Re: Integrals

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.

13. Jul 23, 2012

klondike

Re: Integrals

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.

14. Jul 23, 2012

XodoX

Re: Integrals

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

And

d/dx 2xsin(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.

Last edited: Jul 23, 2012
15. Jul 23, 2012

Robert1986

Re: Integrals

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.

16. Jul 23, 2012

XodoX

Re: Integrals

17. Jul 23, 2012

Robert1986

Re: Integrals

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.

18. Jul 23, 2012

Staff: Mentor

Re: Integrals

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.

19. Jul 23, 2012

XodoX

Re: Integrals

Not really.

20. Jul 23, 2012

Staff: Mentor

Re: Integrals

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.