Question

Math question in the comments.

Answer 1

\[\int\limits_{}^{} (4-x)/(\sqrt{4-x^2}\]

Answer 2

I have it separated into \[\int\limits_{}^{} 4/(\sqrt{4-x^2}) + \int\limits_{}^{}-x/\sqrt{4-x^2}\]

Answer 3

My question is do I leave the 4 where it is? or should i pull it through the integral before trying to solve.

Answer 4

I'd put it through. What grade are you in? I'm just learning this. and i though first you'd want to get rid of the exponent before you doo anything else?

Answer 5

So
\[4*\int\limits_{}^{} 1/\sqrt{4-x^2} + \int\limits_{}^{} -x/\sqrt{4-x^2}\]

Answer 6

I know where to go from this point, but I wonder if not pulling the 4 through the integral would have caused an issue.

Answer 7

Have you tried to solve the problem if you did that? o_o

Answer 8

Yea the u substitution gets weird and i end up with a coefficient of 1/4 instead of 4. I could be making an error tho.

Answer 9

I wouldn't do it that way. Have you learned trig subs yet?

Answer 10

Yea i know the trig subs.

Answer 11

I would do a trig sub.

Answer 12

I got to the correct answer, but honestly i dont know what im doing sometimes.

Answer 13

Hm, not sure how you would get there without a trig sub, but it is possible. :-)

Answer 14

My bad. I did use trig subs to get the answer. I'm just geting 2 different answers based on whether i pull the 4 out or not.

Answer 15

Thats whats confusing me.

Answer 16

I would make x=2sin(u)

Answer 17

it is?

Answer 18

Oh sorry, my bad. It is right.

Answer 19

Not sure why you are getting different answers without being able to see your work.

Answer 20

i would rewrite 4 as 2^2 -x^2
take out the 4 is a good idea
now as you can see \[\frac{ 1 }{ \sqrt{a^2 -x^2} }\]
looks like derivative of some invs trig ratios

Answer 21

oh nvm

Answer 22

\[\int\limits_{}^{} \frac{ 4 }{ \sqrt{4-x^2} }\] i thought this is the original question

Answer 23

I wish it was :/

Answer 24

you can't take out 4 from (4-x) that is illegal lol

Answer 25

I'm not sure what the question really is...
can you post your work?

Answer 26

Im not exactly pulling a 4 out of (4-x)

Answer 27

Sure one second

Answer 28

i see.. i didn't see your 2nd comment

Answer 29

Did you get this for your solution: http://www.wolframalpha.com/input/?i=%284-x%29%2F%284-x^2%29^%281%2F2%29+dx

Answer 30

is your question this:
\[\int\limits \frac{4}{ \sqrt{4-x^2}} dx=4 \int\limits \frac{1}{\sqrt{4-x^2}} dx \\ =4 \int\limits \frac{1}{\sqrt{4-4 \cdot \frac{1}{4} x^2}} dx \\ =4 \int\limits \frac{1}{\sqrt{4(1-\frac{1}{4 } x^2})} dx \\ =4 \int\limits \frac{1}{\sqrt{4} \sqrt{1-\frac{1}{4} x^2}} dx \\ =4 \int\limits \frac{1}{2 \sqrt{1-(\frac{x}{2})^2}} dx \\ =\frac{4}{2} \int\limits \frac{1}{\sqrt{1-(\frac{x}{2})^2}} dx \\ =2 \int\limits \frac{1}{\sqrt{1-(\frac{x}{2})^2}} dx\]

the question is about the 4 in the root? or the 4 on top ? or something else?

Answer 31

du = -2xdx typo

Answer 32

@freckles that was not my question sorry

Answer 33

\[\int\limits_{}^{} \frac{ 4-x }{ \sqrt{4-x^2} }\] this is the question @freckles

Answer 34

Actually theres an error

Answer 35

yes i understand that is the whole initial question
but I want to know his question about the question

Answer 36

I still got to integrate half of that last part

Answer 37

My question is that if you pull a 4 through an integral will it change the answer compared to if you left the 4 in the integral.

Answer 38

are you trying to factored a 4 from the top or from within the root?

Answer 39

I may be doing some incorrect work but i get 2 answers.

Answer 40

4 by itself in the numerator

Answer 41

can you just pull that through the integral?

Answer 42

\[4-x=4(1-\frac{1}{4} x)\]

Answer 43

so thats a yes im assuming

Answer 44

you can do that
but I would just write the integral as two separate integrals first

Answer 45

\(\color{blue}{\text{Originally Posted by}}\) @sweetburger
So
\[4*\int\limits_{}^{} 1/\sqrt{4-x^2} + \int\limits_{}^{} -x/\sqrt{4-x^2}\]
\(\color{blue}{\text{End of Quote}}\)
\[4*\int\limits_{}^{} 1/\sqrt{4-x^2} + \int\limits_{}^{} -x/\color{Red}{4}\sqrt{4-x^2}\]
so there should be 4 at the denominator maybe that's the mistake

Answer 46

\(\color{blue}{\text{Originally Posted by}}\) @freckles
\[4-x=4(1-\frac{1}{4} x)\]
\(\color{blue}{\text{End of Quote}}\)
according to this XD just guessing

Answer 47

you can do this but it looks more dramatic to me than it should
\[4 [ \int\limits \frac{1}{ \sqrt{4-x^2}} dx+\int\limits \frac{-x}{4 \sqrt{4-x^2}} dx]\]

Answer 48

So you have to pull the 4 out of everything.

Answer 49

why not just do this
\[4 \int\limits \frac{1}{\sqrt{4-x^2}} dx-\int\limits \frac{x}{\sqrt{4-x^2}} dx\]

Answer 50

Oh that is what i did.

Answer 51

oh I thought you were talking about pulling a 4 out of the top

Answer 52

well if thats correct I guess i know its possible.

Answer 53

you can always bring constant multiples outside the integral

Answer 54

Thats what i was wondering if the 4 could be taking out like that.

Answer 55

\[\int\limits c f(x) dx=c \int\limits f(x) dx\]

Answer 56

I was going off the rules with summations where you can pull constants through didnt know if it applied here.

Answer 57

Thinking that integrals are kinda related to summations.

Answer 58

they are related

Answer 59

that looked like Riemanns

Answer 60

\[\int\limits\limits_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \Delta x) \Delta x \\ \text{ where } \Delta x =\frac{b-a}{n}\]

Answer 61

integrals are just sums of infinitely many areas of rectangles

Answer 62

Alright makes sense. Thanks for the help @freckles .

Answer 63

well some of those area could be negative depending where the rectangle is falling (like underneath the x-axis)

Answer 64

though area isn't really a negative thing :p

Answer 65

Isnt that why the area for some definite integrals can end up being 0 as they will cancel out?

Answer 66

well it is actually call net area if you have any rectangles below the x-axis
or if your curve goes below the x-axis

Answer 67

Oh didnt know that. Don't think my teacher ever mentioned it.

Answer 68

http://tutorial.math.lamar.edu/Classes/CalcI/AreaProblem.aspx

Answer 69

"In cases where the function is both above and below the x-axis the technique given in the section will give the net area between the function and the x-axis with areas below the x-axis negative and areas above the x-axis positive. So, if the net area is negative then there is more area under the x-axis than above while a positive net area will mean that more of the area is above the x-axis.
"

Answer 70

that was from the bottom of that page

Answer 71

just in case you wanted an example of net area

Answer 72

Alright thanks. I have my final in about an hour but i will take a look at it after.

Answer 73

oh wow
good luck

Answer 74

good luck!!!
mine is tomorrow (so let me know what was on the final) just kidding:P
good luck!!

Answer 75

Thanks again much appreciated. :))