Question

Find the area bounded by the graph x/(3+x^2) and 1x/7

Answer 1

Okay first you have to find where they intersect then you have to take the definite integral of the one minus the definite integral of the other I think.

Answer 2

http://www.wolframalpha.com/input/?i=+the+area+bounded+by+the+graph+x%2F%283%2Bx%5E2%29+and+1x%2F7

They intersect at 0 it seems.

Answer 3

You have to find two intersections one sec let me just make sure I am giving you the right information have you done this type of problem before?

Answer 4

Nope, that is why I'm asking lol.

Answer 5

Okay I think what I said was right, but I have only done a problem similar to this once before. First we should set the equations equal to each other and then get everything on the same side set the other side to zero and find the zeros of the function. That will be the points of intersection.

Answer 6

x/(3+x^2) = 1x/7
(x/(3+x^2)) - x/7 = 0
(7x - 3x - x^3)/(21+7x^2) = 0
Okay so we know for sure x = 0 is an intersection and the other one we find by factoring the top
(x(x-2)(x+2)) so our other solutions are 2 and -2 we only need to worry about on of them because once we find the area between the one 0 and either of the other intersections we can just multiply that by 2 to get the complete answer I think.

Answer 7

one**

Answer 8

Follow so far? I just checked my answer so are intersections so if you are good all we need to do now is definite integrals and subtraction of the top function by the bottom function I mean the definite integrals. Know what I mean?

Answer 9

yessir.

Answer 10

do you know how to integrate this or do you want me to try doing it?

Answer 11

Sorry I mean integrate them you have to integrate them separately not the combined function

Answer 12

Okay I think the first one is (1/2)ln(3+x^2) and the seconds one is (1/14)x^2, but I am not real good with integration yet so these could be wrong one second I gotta check my answers

Answer 13

(7x - 3x - x^3) and (21+7x^2)?

Answer 14

I'm good on the integration part, just not good with the calc 1 stuff haha.

Answer 15

no no you integrate the functions you started with sorry I worded that really poorly

Answer 16

ok np

Answer 17

\[\int\limits\frac{x}{x^2+3}\] u = x^2 + 3
du = 2xdx
1/2 du = xdx
\[1/2 \int\limits \frac{du}{u}\]

Answer 18

= 1/2 ln|u| +c

Answer 19

the next one is a line... y = x/7

Answer 20

if we integrate that then it's x^2/14

Answer 21

yeah sorry I forgot to write plus c yes both of those are correct

Answer 22

So now you need to do the definite integral of the top function from 0 to 2 - the definite integral of the bottom function from 0 to 2 are you good with that?

Answer 23

so (ln(2^2 + 3) - ln(3)) - ((2^2/14) - 0)

Answer 24

looks good =)

Answer 25

http://www.wolframalpha.com/input/?i=%28ln%282%5E2+%2B+3%29+-+ln%283%29%29+-+%28%282%5E2%2F14%29+-+0%29

Answer 26

so i guess the area is that decimal # then...

Answer 27

or the log

Answer 28

Wait you are done you have to multiply by two because there is -2 to 0 and 0 to negative 2 know what I mean?

Answer 29

huh?

Answer 30

I thought it was 0 to 2?

Answer 31

Remember when we factored the zeros were at -2, 0, and 2 but the area between -2 and 0 is symmetrical to the area between 0 and 2 of the curves so all you need to do is multiply the value you have now by 2 and you are done

Answer 32

oh okay yeah.

Answer 33

awesome =) have any more?

Answer 34

apparently that's not correct hmm :(.

Answer 35

http://www.wolframalpha.com/input/?i=2+*%28ln%282%5E2+%2B+3%29+-+ln%283%29%29+-+%28%282%5E2%2F14%29+-+0%29

Answer 36

that one

Answer 37

hmph

Answer 38

hmm one second

Answer 39

how can you tell it is not correct?

Answer 40

I checked the answer in the back of the book.

Answer 41

what was the answer?

Answer 42

Oh wells, getting alittle tired, maybe I'll just continue tomorrow :).

Answer 43

wait could you just give me the actual answer so I could figure out for my own sake haha

Answer 44

Actually... The book might be wrong itself so I'm going to try it out again tomorrow, it's probably the right answer lol stupid textbooks.. Professors write the answers from their students answers and put them in thge book :(.

Answer 45

Alright, but could you just humour me haha I would really appreciate it.