Question

sketch and then calculate the area of the region bounded by the graphs of x=1, y=x+1 and y=1/sqrt(x^2+1)

Answer 1

Is it possible to sketch the graph \(y = \frac{1}{\sqrt{x^2 + 1}}\)? \[\]

Answer 2

I think I would use wolfram alpha to sketch the graph.

Answer 3

Such a complicated Graph to sketch all by yourself.

Answer 4

Okay I got it

We need to fins the intersection of \(\frac{1}{\sqrt{x^2+1}}=y \) and y = x + 1
\[x + 1 = \frac{1}{\sqrt{x^2+1}} \implies (x+1)^2 (x^2+1) = 1\]
To do it quickly I will use wolfram alpha but you should solve on your own. BTW from the graph it is clearly visible that the point is x =0 and y=1

So here is the equation you should use to get the area of the region required

\[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx\]

Answer 5

I did all the talking and solving :-/

Answer 6

sorry somthing came up

Answer 7

i appreciate your help

Answer 8

thats what i tried to do

Answer 9

So, did you get it? I mean the answer did you get the answer?

Answer 10

i got (x^2/2)+x - Lnsec theta+ tan theta

Answer 11

.618626 area?

Answer 12

\[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx = \left[ \frac{x^2}{2} + x \right]_0^1 - \left[ \tan^{-1}x \right]_0^1 \]
I think this is what you are supposed to get.

Answer 13

\[\frac{1}{2} + 1 - \frac{\pi}{4}\]

Answer 14

I am getting .714... something

Answer 15

the second part i got LN(sec theta + tan theta) for integral

Answer 16

did i do it wrong?

Answer 17

Yeah I don't seem to get ln(sec theta) anywhere. You should recheck your solution and I will check mine. Maybe I am wrong.

Answer 18

i let x = tan theta and du = sec^2 theta

Answer 19

did it this way http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%5E2%2B1%29

Answer 20

Umm Lammy I think \[\frac{1}{\sqrt{x^2+1}} = \tan^{-1}x\]

I am not sure.

I am gonna apply same substitution as yours \(dx = d\theta \sec^2\theta\)
\[\frac{1}{\sqrt{\tan^2\theta + 1}}\times \sec^2\theta d\theta = \sec \theta d\theta \]
I think you are right I was wrong, sorry for misleading you

Answer 21

so integral of sec theta = LN abs(sec theta + tan theta) right?

Answer 22

no no you were helping =)

Answer 23

then i just subtract and will get the answer right?

Answer 24

yeah

Answer 25

do you think you have time to help me with on more? befor you go to sleep?

Answer 26

hahah

Answer 27

yeah sure, post it :-D

Answer 28

just posted it

Answer 29

=)