Integration (Advanced Higher)

**shuggyboi** · October 10th 2007, 12:52 PM

Hey,

Im new here so firstly - Hello to everyone and i look forward to being part of this community

Second, my problem:

Thanks in advance,
Shug

p.s. the weird sign at the end of sin is a theta sign.

**Jhevon** · October 10th 2007, 12:57 PM

Originally Posted by shuggyboi

Hey,

Im new here so firstly - Hello to everyone and i look forward to being part of this community

Second, my problem:

Thanks in advance,
Shug

p.s. the weird sign at the end of sin is a theta sign.

you have it, why are you stuck? do you understand what you did so far, or did you just copy this from someone? if $x = \sin \theta$ then, $dx = \cos \theta ~d \theta$ and,

$\int_{0}^{1}\sqrt{1 - x^2}~dx = \int_{0}^{\frac {\pi}2} \sqrt{1 - \sin^2 \theta}~\cos \theta d \theta = \int_{0}^{\frac {\pi}2} \cos^2 \theta~d \theta$

now continue

**tukeywilliams** · October 10th 2007, 12:57 PM

$\int_{0}^{1} \sqrt{1-x^2} \ dx$ .

$x = \sin \theta$

$dx = \cos \theta \ d \theta$

So we have $\int_{0}^{\frac{\pi}{2}} \cos^{2} \theta \ d \theta$ .

**shuggyboi** · October 10th 2007, 01:03 PM

ahhh!!!

thanks.. sorry

i was making a stupid mistake..

i totally forgot about the root sign :|

many thanks

**angel.white** · December 4th 2007, 10:28 AM

Okay, I was looking at this forever last night, and finally gave up. Today, I went to the math lab (basically math tutoring) to ask those guys to do it. As soon as I point to the problem in my book, the answer just comes to me >.<

I've attached a graph to illustrate this (I would have figured it out from the graph last night, but my problem was to integrate x+(1-x^2)^.5 so the graph looked different, and I didn't realize what I really had)

So, to save others the same frustration I had:

Origional integral
$\int_{0}^{1} \sqrt{1-x^{2}} dx$

Set the function being integrated as equal to y
$y=\sqrt{1-x^{2}}$

Realize that y must be positive, because no real number squared will give you a negative number.
$y\geq 0$

Square the function
$y^{2}=1-x^{2}$

Add x^2 to both sides
$y^{2}+x^{2}=1$

Now realize that this is the equation of a circle. And because y must be positive, it is the equation for the top half of a circle.

Now we are trying to find the area from 0 to 1 under the top half of a circle, and who's center is the origin. Because the center is the origin, we know the area is only what is inside of the circle (ie if it were raised 1 unit, we would have to account for the additional area, but in this case we do not). And because it is the top half of the circle with radius 1 from 0 to 1, then we are taking the right half of the top half of the circle, so just a quarter of the circle who's radius is 1.

Area of a circle is $\pi r^{2}$ and r = 1, so the area of our circle is $\pi$ . And since a quarter of the circle will have a quarter of the area, we can simply take a quarter of the total area. $\frac{\pi}{4}$

Since integrals calculate the area under the graph of a function, from a to b, we are trying to find the area under $\sqrt{1-x^{2}}$ and since $\frac{\pi}{4}$ is that area, then $\frac{\pi}{4}$ must be the solution to the integral.

Thus:
$\int_{0}^{1} \sqrt{1-x^{2}} dx = \frac{\pi}{4}$

Image attached to illustrate:

**Jhevon** · December 4th 2007, 10:56 AM

Originally Posted by angel.white

Okay, I was looking at this forever last night, and finally gave up. Today, I went to the math lab (basically math tutoring) to ask those guys to do it. As soon as I point to the problem in my book, the answer just comes to me >.<

I've attached a graph to illustrate this (I would have figured it out from the graph last night, but my problem was to integrate x+(1-x^2)^.5 so the graph looked different, and I didn't realize what I really had)

So, to save others the same frustration I had:

Origional integral
$\int_{0}^{1} \sqrt{1-x^{2}} dx$

Set the function being integrated as equal to y
$y=\sqrt{1-x^{2}}$

Realize that y must be positive, because no real number squared will give you a negative number.
$y\geq 0$

Square the function
$y^{2}=1-x^{2}$

Add x^2 to both sides
$y^{2}+x^{2}=1$

Now realize that this is the equation of a circle. And because y must be positive, it is the equation for the top half of a circle.

Now we are trying to find the area from 0 to 1 under the top half of a circle, and who's center is the origin. Because the center is the origin, we know the area is only what is inside of the circle (ie if it were raised 1 unit, we would have to account for the additional area, but in this case we do not). And because it is the top half of the circle with radius 1 from 0 to 1, then we are taking the right half of the top half of the circle, so just a quarter of the circle who's radius is 1.

Area of a circle is $\pi r^{2}$ and r = 1, so the area of our circle is $\pi$ . And since a quarter of the circle will have a quarter of the area, we can simply take a quarter of the total area. $\frac{\pi}{4}$

Since integrals calculate the area under the graph of a function, from a to b, we are trying to find the area under $\sqrt{1-x^{2}}$ and since $\frac{\pi}{4}$ is that area, then $\frac{\pi}{4}$ must be the solution to the integral.

Thus:
$\int_{0}^{1} \sqrt{1-x^{2}} dx = \frac{\pi}{4}$

Image attached to illustrate:

Yes, that is an excellent solution. A nice way to do the problem without integration. You should remember what you did here. Next time you shouldn't have to go through the algebraic manipulation to get the circle, you would immediately know that $y = \sqrt{1 - x^2}$ represents the upper half of the unit circle while $y = - \sqrt{1 - x^2}$ represents the lower half of the unit circle.

and in general, $y = \pm \sqrt{r^2 - x^2}$ , where $r \in \mathbb{R}$ , represents the upper and lower halves, respectively, of the circle with center $(0,0)$ and radius $r$ .

What would be something like: $\int_0^4 \sqrt{4 - (x - 2)^2}~dx$ ?

**angel.white** · December 4th 2007, 11:10 AM

Originally Posted by Jhevon

What would be something like: $\int_0^4 \sqrt{4 - (x - 2)^2}~dx$ ?

Well the x-2 means the center will be (2,0)
and the 4 means the radius will be 2

So the circle will span the domain of [0,4]

So $\pi r^{2}$ becomes $4\pi$

And since this is the upper half of a circle, then half of the total area is $2\pi$

So
$\int_0^4 \sqrt{4 - (x - 2)^2}~dx=2\pi$

On a side note, when you wrote the integral, you placed a tilde before "dx" I'm curious what this does/means.

**Jhevon** · December 4th 2007, 11:15 AM

Originally Posted by angel.white

Well the x-2 means the center will be (2,0)
and the 4 means the radius will be 2

So the circle will span the domain of [0,4]

So $\pi r^{2}$ becomes $4\pi$

And since this is the upper half of a circle, then half of the total area is $2\pi$

So
$\int_0^4 \sqrt{4 - (x - 2)^2}~dx=2\pi$

Yup! You're a genius!

On a side note, when you wrote the integral, you placed a tilde before "dx" I'm curious what this does/means.

It is just used to give a space. some users use \ or \, or \; but i find all of those unaesthetic. I want both my LaTex and my LaTex code aesthetic. Math is half about being elegant after all.

without the ~: $\int_0^4 \sqrt{4 - (x - 2)^2}dx$

with the ~: $\int_0^4 \sqrt{4 - (x - 2)^2}~dx$

you see how the dx is not stuck on the integrand? they give each other space

**topsquark** · December 4th 2007, 11:17 AM

Originally Posted by angel.white

On a side note, when you wrote the integral, you placed a tilde before "dx" I'm curious what this does/means.

It adds a space. For an integral like this one, I find that the space makes it look nicer.

$\int_0^4 \sqrt{4 - (x - 2)^2}~dx=2\pi$
vs
$\int_0^4 \sqrt{4 - (x - 2)^2}dx=2\pi$

-Dan

**angel.white** · December 4th 2007, 12:39 PM

Aah, that makes sense. I don't like how it effects searches, though, makes it harder to search for things because you have to try various different ways of writing a function, and say someone already has solved the problem you're working on, but they wrote it a certain way, and you just can't find the right syntax to plug into the search feature, so you never find the thread. A trade off, I guess.

I think the site would really benefit from integrating the LaTeX into the search feature. Like a way to search through [tex] tags, and return places they had been used with links to the actual posts. Because right now, it's only easy to search for story problems, because they usually use the same phrases word for word. But if you have just an equation and are told to "differentiate" or "integrate" and you go to search for it, you have to hit the syntax just right to find the answer, and frequently instructors will change constants in order to reuse the same problem, so your answer might be out there, but you never find it because you are looking for $\frac{2}{3}x^{2}$ instead of $\frac{3}{4}x^{2}$ .

Anyway, it would be helpful because it would reduce redundant problems, and allow users to capitalize off of the previously answered questions.

**Krizalid** · December 4th 2007, 12:41 PM

Geometric reasoning is the best way to get rid off this.

First thought: we're integratin' an even function, so rewrite

$\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = 2\int_0^1 {\sqrt {1 - x^2 } \,dx} .$

We have a half circle of radius 1, so the area will be $\frac\pi2,$

$2\int_0^1 {\sqrt {1 - x^2 } \,dx} = \frac{\pi }<br /> {2}\,\therefore \,\int_0^1 {\sqrt {1 - x^2 } \,dx} = \frac{\pi }<br /> {4}.$

--

I prefer using \, - the space looks better.

**Jhevon** · December 4th 2007, 12:58 PM

Originally Posted by angel.white

Aah, that makes sense. I don't like how it effects searches, though, makes it harder to search for things because you have to try various different ways of writing a function, and say someone already has solved the problem you're working on, but they wrote it a certain way, and you just can't find the right syntax to plug into the search feature, so you never find the thread. A trade off, I guess.

I think the site would really benefit from integrating the LaTeX into the search feature. Like a way to search through [tex] tags, and return places they had been used with links to the actual posts. Because right now, it's only easy to search for story problems, because they usually use the same phrases word for word. But if you have just an equation and are told to "differentiate" or "integrate" and you go to search for it, you have to hit the syntax just right to find the answer, and frequently instructors will change constants in order to reuse the same problem, so your answer might be out there, but you never find it because you are looking for $\frac{2}{3}x^{2}$ instead of $\frac{3}{4}x^{2}$ .

Anyway, it would be helpful because it would reduce redundant problems, and allow users to capitalize off of the previously answered questions.

right you are. but it would be really hard to install such a feature, since we can only search by the code used. and besides using things like ~ or \, users type there codes differently. some users don't use spaces in code, while i do (i think it makes the code look nicer, which as i said, is important to me for some reason). so i would type \cos^2 \left( \frac {\sqrt { x^3 + 3 }}4 \right) while another types \cos^2\left(\frac{\sqrt{ x^3 + 3 }}4\right) to get $\cos^2 \left( \frac {\sqrt { x^3 + 3 }}4 \right)$

a feature that can recognize that both these codes are saying the same thing would have to have A.I. or something

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