# Hard integral problem

#### 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:

#### whozum

Hint: fundamental theorem of calculus..

#### Corneo

If I recall the FTC then it should be

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

#### 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.

#### 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:

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

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

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#### infirmus

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

Ahh well. its all good. Thanks.

#### dextercioby

Homework Helper
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.

#### infirmus

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

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

Daniel.

#### saltydog

Homework Helper
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:

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

#### amcavoy

saltydog said:
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:

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

#### saltydog

Homework Helper
alexmcavoy@gmail.com said:
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?
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.

#### 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

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#### saltydog

Homework Helper
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:

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

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
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:

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

#### 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.

#### whozum

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

#### saltydog

Homework Helper
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 $$\frac{dH}{dx}$$ 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:

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

#### amcavoy

saltydog said:
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:

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

Homework Helper
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:

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

#### amcavoy

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

#### GCT

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

$$e^{it^{2}}~=~cost^{2}+isint^{2}~,~e^{-it^{2}}~=~cost^{2}-isint^{2}$$

$$e^{it^{2}}-e^{-it^{2}}=2cost^{2}~,~(e^{it^{2}}-e^{-it{2}})/2=cost^{2}}$$

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