Question

K im having trouble using the integral to find the surface are of y=sqrt(1+4x), 1 13 years ago

Answer 1

again :)...you're certainly in hell !

Answer 2

ya, im taking cal 2 over the summer class started on monday first test 2morrow

Answer 3

\[\int\limits_{1}^{5}f(x)\sqrt{1+ f \prime^2(x) }dx\]

Answer 4

im pretty lost on this one i got (dy/dx)^2 = 4/(1+4x)

Answer 5

and that might be my problem

Answer 6

good luck then ! I've forgotten 2 Pi !

Answer 7

thank you and ya I know the formulas lol its using them that gets me

Answer 8

ya if you can just give me a shove into the integral part i think i can take it from there

Answer 9

yeah..\[\int\limits_{1}^{5}\sqrt{1+4x}*\sqrt{1+4/(1+4x)}dx=\int\limits_{1}^{5}\sqrt{1+4x+4}dx\] you can finish it now ?

Answer 10

well I see what you did kind of.. but how exactly did you do that? cross multiply?!?!

Answer 11

did you foil the nominators and that canceled the denominator. haha mind blown

Answer 12

\[\sqrt{a}\sqrt{b}=\sqrt{ab}\]

Answer 13

and the one on the left is the same thing as sqrt(1+4x)/1 right

Answer 14

sorry ; I have problem with my connection ! I couldn't post quickly ! aaargh..it takes minutes !!!!

Answer 15

give me the final answer ! so I can check !

Answer 16

so Im still kind of confused

\[\sqrt{a}\times \sqrt{b/1+4x} = \sqrt{(a(b))/1+4x}\]

Answer 17

kk! step by step !

Answer 18

ya, sorry just this part is throwing me for a loop

Answer 19

\[\sqrt{1+4x}\sqrt{1+4/(4+x)}=\sqrt{(1+4x)1+(1+4x)4/(1+4x)} \]

Answer 20

(1+4x)4/(1+4x)=4

Answer 21

no problem ! I'm here for help !

Answer 22

ok so look at the first message you posted after my sorry message lol the 4 after (1+4x)*4* is that suppose to be there?

Answer 23

k never mind i got it haha sorry

Answer 24

so the final answer is.....?!

Answer 25

k so i got pi/3(5^(3/2) -1)

Answer 26

does that look good?

Answer 27

hang on i forgot to sub back in

Answer 28

sweet! i got 98pi/3 and thats the answer!!!

Answer 29

no :( !I checked it 5 times...

Answer 30

yeaaah now right !

Answer 31

no to 98pi/3

Answer 32

oh lol ok

Answer 33

sweet well be on the look out im bound to post a couple more

Answer 34

w8 w8 !

Answer 35

actually while i got you the derivative of y=sin(pi(x)) is (picos(pi(x))) right

Answer 36

seems to be smthing wrong ....Just send me the link kk...

Answer 37

to my question?

Answer 38

no your answer !

Answer 39

and yes...pi cos(pi(x)....Aarghh connection makes me mad !

Answer 40

ok so how would I square that? would it be pi cos^2(pi(x))

Answer 41

sorryy !!

Answer 42

its kewl

Answer 43

anything els! and you were right 98/3 pi

Answer 44

right ya how would I square the pi cos(pi(x)) to use it in the same formula? would it just be pi cos^2(pi(x))

Answer 45

pi^2 no ?!

Answer 46

so it would be pi^2 cos^2(pi(x))

Answer 47

yeaaah !

Answer 48

(ab)^2=a^2b^2

Answer 49

ok so for this problem its the same thing just with this function

y=sin(pi(x))

I've got this

Answer 50

\[\int\limits_{0}^{1} 2\pi \sin(\pi(x))\]

Answer 51

oops one sec

Answer 52

\[\int\limits_{0}^{1} 2\pi \sin(\pi(x))\sqrt{1+(\pi^2 \cos^2(\pi(x))}\]

Answer 53

k there we go

Answer 54

yeah ! what do you suggest ?

Answer 55

really I have no Idea it looks super messy. all I can think of is u substitution put it looks like it will need a substitution within a substitution. I'm hoping not to have to mess with that

Answer 56

this one is not easy !

Answer 57

trig is always a problem!

Answer 58

are you figuring it out man?

Answer 59

I have an Idea..but I'm not sure yet....cos(x)=u !

Answer 60

variabl change !

Answer 61

cos(pix) =u sorry !

Answer 62

cos(pix)=u It really works ; but I'll give you a better solution !
let G be \[G=\int\limits_{}^{}\sqrt{1+\pi^2u^2}du\]
so ! obviously \[G \prime(u)=\sqrt{1+\pi^2u^2} \]
then,\[S=2\pi \int\limits_{0}^{1}\sin(\pi x)G \prime(\cos(\pi x))dx=-2\int\limits_{0}^{1}(\cos(\pi x))'G'(\cos(\pi x)dx\]
wich equals \[S=-2\left[ \left[ G(\cos(\pi x)\right] \right]\] from 0 to 1
so let's find G using integration by parts
\[G=\int\limits_{}^{}(u)'\sqrt{1+\pi^2u^2}du=u \sqrt{1+\pi^2u^2}- \int\limits_{}^{}\pi^2u^2/(\sqrt{1+\pi^2u^2})\]
a little tricky right here ! to calculate the second part we add 0=+1-1
\[\int\limits\limits_{}^{}\ (\pi^2u^2+1-1)/(\sqrt{1+\pi^2u^2})du=\int\limits\limits_{}^{}\ (\pi^2u^2+1)/(\sqrt{1+\pi^2u^2})du-\int\limits\limits_{}^{}\ 1/(\sqrt{1+\pi^2u^2})du\]
let's begin with the second one !
\[\int\limits_{}^{}1/\sqrt{1+\pi^2u^2}du=1/\pi(\ln \left| u+\sqrt{u^2+\pi^-2} \right|)\]
now the firt one !\[\int\limits\limits\limits_{}^{}\ (\pi^2u^2+1)/(\sqrt{1+\pi^2u^2})du =\int\limits_{}^{}\sqrt{1+\pi^2u^2}du=G\]
let now put those back to * *
\[G=u \sqrt{1+\pi^2u^2}-G-1/\pi(\ln \left| u+\sqrt{u^2+\pi^-2} \right|)\\]
\[2G=u \sqrt{1+\pi^2u^2}-1/\pi(\ln \left| u+\sqrt{u^2+\pi^-2} \right|)\ \]

Answer 63

so \[S=2\pi(G(1)-G(0))\]
You owe me a beer ! bad girl !