Hard integral problem

1. Jun 19, 2005

infirmus

Can someone give me a hand with this problem - Its screwing with my head!

The derivative of the function $f(x) = \int_{43}^{x^3}\cos(t^2){dt}$ is:

2. Jun 19, 2005

whozum

Hint: fundamental theorem of calculus..

3. Jun 19, 2005

Corneo

If I recall the FTC then it should be

$$f'(x) = 3x^2 \cos x^6$$

4. Jun 19, 2005

infirmus

Dam dont even remember what FTC is... I remember hearing bout it..

I guess I'll go read up my text book.

Corneo - thats the correct answer...

Thanks guys, great help.

5. Jun 19, 2005

whozum

Fundamental theorem of calculus says:

$$\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$$

So applying it to this problem, we have a = 43, and x = x^3
$$\frac{d}{dx} \int_{43}^{x^3} \cos t^2 dt = f(x^3)$$

Since the upper limit is a function of x, the chain rule is applied. The FTOC with the chain rule is:

$$\frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x))g'(x)$$

$$\frac{d}{dx} \int_{43}^{x^3} \cos t^2 dt = \cos \left((x^3)^2\right) 3x^2 = 3x^2\cos (x^6)$$

Last edited: Jun 19, 2005
6. Jun 19, 2005

infirmus

Hmm, I feel really stupid... :shy:

Ahh well. its all good. Thanks.

7. Jun 19, 2005

dextercioby

Whozum,you're completely incorrect.

Denoting the antiderivative of f(t) by F(t),what's this equal to ?

$$\frac{d}{dx}\left[ F(t)\left|_{t=a}\right \right]$$

According to you,it's f(a).According to me,it's 0.

So what's the constant doing in the last line...?Was it there a linear term in "x" before differentiation...?

Daniel.

8. Jun 19, 2005

infirmus

I wouldn't say that it was completely incorrect. A bit hard to understand maybe, but I got the gist of it... Not using capital F's for integrals didnt help. He did say:
F(43) is a constant, and so its derivative is 0 and can be ommited.

9. Jun 19, 2005

dextercioby

All his 4 formulae (written in LaTex) are incorrect.So he was completely incorrect.

Daniel.

10. Jun 19, 2005

saltydog

Tell you what Infirmus, check out Leibnitz rule and kindly solve the following:

$$\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt$$

You can do it I bet! Just follow the rules and do all the differentiations. The answer is impressive. And after that, just come up with some examples for the functions and do some real ones, say:

$$\frac{d}{dx} \int_{x^2}^{e^x} (x^2t+xtSin[x]) dt$$

11. Jun 19, 2005

amcavoy

It seems to me that $$\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt$$ would be $$f(v(x),t)\frac{\partial v}{\partial x}-f(u(x),t)\frac{\partial u}{\partial x}$$ but that isn't the answer, is it?

12. Jun 19, 2005

saltydog

No Alex, that's not quite right. Remember, the integration is being done with respect to t and so at some point, you'll get a function in t that you then substitute v(x) and u(x) where ever t occurs. So if anything, it would include expressions such as:

$$f(x,u(x))$$

and:

$$f(x,v(x))$$

right?

But there are some other components also.

13. Jun 19, 2005

amcavoy

Right, I don't know why I did that. Anyways, I can understand that part, it's just that I can't get past why there is this:

$$\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt$$

I think this is the answer:

$$\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt+f(x,v(x))\frac{\partial v}{\partial x}-f(x,u(x))\frac{\partial u}{\partial x}$$

according to MathWorld.

Any ideas?

Thanks,

--Alex

Last edited: Jun 19, 2005
14. Jun 19, 2005

saltydog

Very good Alex. Suppose we have:

$$H(u,v,x)=u^2+2uv+x$$

with:

$$u=f(x);\quad v=g(x)$$

Then:

$$\frac{dH}{dx}=\frac{\partial H}{\partial u} \frac{du}{dx}+\frac{\partial H}{\partial v} \frac{dv}{dx}+\frac{dH}{dx}$$

So let:

$$H(u,v,x)=\int_{v(x)}^{u(x)} f(x,t) dt$$

15. Jun 19, 2005

amcavoy

Alright, I guess I didn't think to write it out that way. So the part I was confused about comes from the $$\frac{dH}{dx}$$ at the end right?

Thanks again,

--Alex.

16. Jun 19, 2005

whozum

I amended the post, is that better for you dex?

17. Jun 19, 2005

saltydog

The final partial treats u and v as constants. Justifications are needed to switch the order of limits but its:

$$\frac{\partial}{\partial x}\int_v^u f(x,t)dt=\int_v^u\mathop\lim\limits_{h\to 0} \frac{f(x+h,t)-f(x,t)}{h} dt=\int_v^u \frac{\partial}{\partial x}f(x,t)dt$$

18. Jun 25, 2005

amcavoy

Sorry to bring it up again, but everything comes out correctly except for the sign when I use the chain rule. I come up with the same thing, but every term is added (just like the chain rule is written). Why is this?

19. Jun 25, 2005

saltydog

Hello Alex:

$$\frac{d}{dx} \int_x^k f(t)dt=-f(x)$$

20. Jun 25, 2005

amcavoy

Ah right. I wasn't treating "u" as a constant. Thanks again.