Hard integral problem

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

The derivative of the function [itex]f(x) = \int_{43}^{x^3}\cos(t^2){dt}[/itex] is:
 
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Hint: fundamental theorem of calculus..
 
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If I recall the FTC then it should be

[tex]f'(x) = 3x^2 \cos x^6 [/tex]
 
Dam dont even remember what FTC is... I remember hearing bout it.. :rolleyes:

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

Corneo - thats the correct answer...

Thanks guys, great help.
 
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Fundamental theorem of calculus says:

[tex] \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) [/tex]

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

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

[tex] \frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x))g'(x)[/tex]

[tex] \frac{d}{dx} \int_{43}^{x^3} \cos t^2 dt = \cos \left((x^3)^2\right) 3x^2 = 3x^2\cos (x^6)[/tex]
 
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Hmm, I feel really stupid... :shy:

Ahh well. its all good. Thanks.
 

dextercioby

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Whozum,you're completely incorrect.

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

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

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.
 
dextercioby said:
Whozum,you're completely incorrect.
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:
whozum said:
Since the upper limit is a function of x, the chain rule is applied. f(43) is omitted since it is a constant.
F(43) is a constant, and so its derivative is 0 and can be ommited.
 

dextercioby

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

Daniel.
 

saltydog

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

[tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex]

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:

[tex]\frac{d}{dx} \int_{x^2}^{e^x} (x^2t+xtSin[x]) dt[/tex]
 
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saltydog said:
Tell you what Infirmus, check out Leibnitz rule and kindly solve the following:

[tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex]

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:

[tex]\frac{d}{dx} \int_{x^2}^{e^x} (x^2t+xtSin[x]) dt[/tex]
It seems to me that [tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex] would be [tex]f(v(x),t)\frac{\partial v}{\partial x}-f(u(x),t)\frac{\partial u}{\partial x}[/tex] but that isn't the answer, is it?
 

saltydog

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alexmcavoy@gmail.com said:
It seems to me that [tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex] would be [tex]f(v(x),t)\frac{\partial v}{\partial x}-f(u(x),t)\frac{\partial u}{\partial x}[/tex] but that isn't the answer, is it?
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:

[tex]f(x,u(x))[/tex]

and:

[tex]f(x,v(x))[/tex]

right?

But there are some other components also.
 
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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:

[tex]\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt[/tex]

added to the answer.

I think this is the answer:

[tex]\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}[/tex]

according to MathWorld.

Any ideas?

Thanks,

--Alex
 
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saltydog

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alexmcavoy@gmail.com said:
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:

[tex]\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt[/tex]

added to the answer.

I think this is the answer:

[tex]\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}[/tex]

according to MathWorld.

Any ideas?

Thanks,

--Alex
Very good Alex. Suppose we have:

[tex]H(u,v,x)=u^2+2uv+x[/tex]

with:

[tex]u=f(x);\quad v=g(x)[/tex]

Then:

[tex]\frac{dH}{dx}=\frac{\partial H}{\partial u} \frac{du}{dx}+\frac{\partial H}{\partial v} \frac{dv}{dx}+\frac{dH}{dx}[/tex]

So let:

[tex]H(u,v,x)=\int_{v(x)}^{u(x)} f(x,t) dt[/tex]
 
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Alright, I guess I didn't think to write it out that way. So the part I was confused about comes from the [tex]\frac{dH}{dx}[/tex] at the end right?

Thanks again,

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

saltydog

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alexmcavoy@gmail.com said:
Alright, I guess I didn't think to write it out that way. So the part I was confused about comes from the [tex]\frac{dH}{dx}[/tex] at the end right?

Thanks again,

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

[tex]\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[/tex]
 
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saltydog said:
Very good Alex. Suppose we have:

[tex]H(u,v,x)=u^2+2uv+x[/tex]

with:

[tex]u=f(x);\quad v=g(x)[/tex]

Then:

[tex]\frac{dH}{dx}=\frac{\partial H}{\partial u} \frac{du}{dx}+\frac{\partial H}{\partial v} \frac{dv}{dx}+\frac{dH}{dx}[/tex]

So let:

[tex]H(u,v,x)=\int_{v(x)}^{u(x)} f(x,t) dt[/tex]
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?
 

saltydog

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alexmcavoy@gmail.com said:
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?
Hello Alex:

[tex]\frac{d}{dx} \int_x^k f(t)dt=-f(x)[/tex]
 
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Ah right. I wasn't treating "u" as a constant. Thanks again.
 

GCT

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not quite sure, but would the following help in solving this integral?

[tex]e^{it^{2}}~=~cost^{2}+isint^{2}~,~e^{-it^{2}}~=~cost^{2}-isint^{2} [/tex]

[tex]e^{it^{2}}-e^{-it^{2}}=2cost^{2}~,~(e^{it^{2}}-e^{-it{2}})/2=cost^{2}} [/tex]
 

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