Integrating cos(u^2): A Calc One Challenge

[whozum]

$$\int cos(u^2)du$$

Is it doable at a Calc One level? I tried by parts and got to

$$\int cos(u^2)du = ucos(u^2) + 2\int(u^2sin(u^2)du$$

but I am having a brain fart as to hwo to advance, trying again by parts.

 

[whozum]

$$y = \int_{3}^{5} cos(u^2)du$$

 

[Data]

Use the fundamental theorem of calculus and the chain rule... you shouldn't have to integrate anything.

 

[whozum]

But were integrating and deriving different variables.
Sorry I don't see what your trying to tell me.

I also kinda need this quick.. its for a friend and its due in 20 minutes

 

[whozum]

$$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$$

$$\frac{d}{dx} \int cos(u^2) du = \left(\frac{d}{du} \int cos(u^2) du \right) \left(\frac{du}{dx}\right)$$

Am I on the right track? The right hand simplifies I know

 

[Data]

Edit: Note I made a little mistake the first time I posted this, it's fixed now though.

ok... since you need this quickly

FTC says

$$f(x) = \frac{d}{dx} \int_a^x f(u) \ du$$

so chain rule says

$$\frac{d}{dx} \int_a^{g(x)} f(u) \ du = f(g(x)) g^\prime(x)$$

and from this you can also work out that

$$\frac{d}{dx} \int_{h(x)}^{g(x)} f(u) \ du = f(g(x))g^\prime(x) - f(h(x))h^\prime(x).$$

Now just figure out what $h, \ f,$ and $g$ are in this case and you're done~

 

[whozum]

Why have I never seen this before? FTC in stewart's book only deals with the derivative of an integral of the same variable.

That makes perfect sense. Thanks a lot.

 

[Data]

Now you have me curious. How does your textbook state it? I've never seen it much differently.

 

[whozum]

I don't have it on me, I'm not at home but I'll let you know.

The FTOC doesn't apply to discontinuous functions, right? If I was tryin to find

$$\int_{-3}^{3} \frac{1}{x^2} dx$$

I would need to use improper integrals

 

[Data]

You would. One of the conditions is that the integrand must be continuous on the (closed) interval between the bounds of integration, so it also has to be bounded on that interval.

Take a look at the mathworld page for details (the second theorem can be generalized a little further than what mathworld states too, but that's not what you would try to use to evaluate that integral).

 

[whozum]

a lot of editting there. :D

I don't get why this professor would give his student that problem when they haven't learned even integration by parts let alone improper integrals and such. He just gave them this one that I don't see how he cna solve without a really messy trig subs (which they also haven't learned).

$$\int_2^{20} \frac{\sqrt{200x-x^2}}{x}dx$$

Even after messing with the square roots I can't get anywhere.

 

[Pyrrhus]

Whozum, it looks to me you are confusing the dummy variable in the integral with the independent variable, or derivative with respect to.

For example

$$\int_{0}^{x} f(t) dt = F(x)$$

x is the real variable, t it's just a dummy variable, it could be as well u.

 

[Data]

it's not a terrible integral. Simplify it to

$$\int \frac{\sqrt{200-x}}{\sqrt{x}} \ dx$$

and then sub $u = \sqrt{200-x}$, and see where that gets you (it should get it to a standard integral).

 

[whozum]

$$du = \frac{-1}{2\sqrt{200-x}}$$

I don't have that in my integral, maple solved it with trig subs too. Messy answer.

 

[whozum]

The question was, if the curve n'(t) = (the above integrand) represents the rate of change of profits, how much moeny was made between week 2 and week 20?

 

[Pyrrhus]

I believe that Data suggested was for

$$u = \sqrt{200 - x}$$

so

$$200 - u^2 = x$$

and

$$- 2udu = dx$$

so

$$\int \frac{-2u^2 du}{\sqrt{200-u^2}}$$

which is more approachable

 

[Data]

Using that sub I can get it to

$$\int \frac{-2u^2}{\sqrt{200-u^2}} \ du$$

which I consider to be a standard integral, though I guess if you haven't seen it enough times in the past it might not seem that way . You can finish it by integration by parts if you need to (though it is a little strange that they wouldn't teach either int. by parts or trig substitution before giving this integral).

 

[Data]

if you're using

$$\int x \ dy = xy - \int y \ dx$$

then let $x = -2u, dy = u\ du/\sqrt{200-u^2}$. You'll also have to make a trigonometric substitution to evaluate one of the resulting integrals.

 

[Hippo]

... maple solved it with trig subs too. Messy answer.

Is it really?

$$\int \frac{\sqrt{200-x}}{\sqrt{x}} \ dx = \int \sqrt{\frac{200}{x}-1} \ dx$$

Let

$$x=200cos^{2}(\theta)$$

Then

$$\int \sqrt{\frac{200}{x}-1} \ dx = \int -400cos(\theta)sin(\theta)\sqrt{sec^{2}(\theta)-1} \ d\theta = \int -400cos(\theta)sin(\theta)tan(\theta) \ d\theta$$

$$= \int -400sin^{2}(\theta) \ d\theta = \int 200cos(2\theta) - 200 \ d\theta = 100sin(2\theta) -200\theta + K$$

and just reevaluate your original limits for the definite inegral.

 

[whozum]

if you're using

$$\int x \ dy = xy - \int y \ dx$$

then let $x = -2u, dy = u\ du/\sqrt{200-u^2}$. You'll also have to make a trigonometric substitution to evaluate one of the resulting integrals.

Thats what I'm saying, it makes no sense to me if theyre being tested on the FTOC to have to solve integrals such as this one, it also emphasizes the resulting answer to be exact. I haven't integrated funky functions much, more standard form ones just to get the jist of what integrating is, but I am sure there's an integral table iwth that form on there. If I don't know it, I'm sure someone who was introduced to integrals a few weeks ago wouldn't either.

Hippo,

I understand, but as you can see above, its expected to be solved by someone who doesn't know integration by parts or trig substitutions.

Maple's answer involved 2 arcsin's and three square roots I believe.

 

[whozum]

Whozum, it looks to me you are confusing the dummy variable in the integral with the independent variable, or derivative with respect to.

For example

$$\int_{0}^{x} f(t) dt = F(x)$$

x is the real variable, t it's just a dummy variable, it could be as well u.

Yeah I'm trying to figure this out in my head, but for some reason it still seems a bit sketchy, I know what you mean though.

 

[Data]

Well, the first part of the FTC is precisely what let's you say (for a sufficiently well-behaved function $f$) that

$$\int_a^b f(x) \ dx = F(b) - F(a)$$

where $F^\prime(x) = f(x)$, is true at all.

 

[whozum]

Is that analogous to

$$\int_a^b f'(x) dx = f(b)-f(a)$$

Because that's all I remember. I just looked at Stewart's book and it is mentioned both ways, I must have just overlooked that one.

 

[Data]

it's exactly the same thing (note my little $F \ ^\prime (x) = f(x)$ at the end ).