Question

Sorry...trouble with integration.
See below.

Answer 1

@calculusfunctions sorry my browser crashed halfway so i can't type it in there. can u give me the link?

Answer 2

If given the expression\[\sqrt{x ^{2}+a ^{2}}\]then let\[x =a \tan \theta\]where\[-\frac{ \pi }{ 2 }<\theta <\frac{ \pi }{ 2 }\]because of trigonometric identity\[1+\tan ^{2}\theta =\sec ^{2}\theta\]Thus if\[\int\limits_{}^{}\sqrt{x ^{2}+3}dx\]then\[x =\sqrt{3}\tan \theta\]and\[dx =\sqrt{3}\sec ^{2}\theta d \theta\]This was where we had left off previously. Now substitute for x and dx to write the integral in terms of θ.

Answer 3

sorry that u had to retype all that. :(

Answer 4

okay so
\[\int\limits_{1}^{3} ((\sqrt{3}\tan \theta)^2+3)(\sqrt{3}\sec^2 \theta) d \theta\]
is that what you mean?

Answer 5

You never mentioned the upper and lower limits on the integral before. Are you now changing the question?

Answer 6

no. i forgot to put them. sorry :{

Answer 7

does it make a differnce?

Answer 8

the directions say to use property 8 which is
if m less than or equal to f(x) less than or equal to M for a less than or equal to xless than or equal to b, then
\[m(b-a) \le \int\limits_{a}^{b}f(x)dx \le M(b-a)\]

Answer 9

Well once we do the trig substitution these will also have to be written in terms of their θ value. Thus if\[x =\sqrt{3}\tan \theta\] then \[\theta =\frac{ \pi }{ 6 }\]if x = 1 and\[\theta = \tan^{-1}\frac{ 2\sqrt{3} }{ 3 }\] if x = 2.

Answer 10

Did you understand? so now we would have\[\int\limits_{1}^{2}\sqrt{x ^{2}+3}dx\]\[=\int\limits\limits\limits_{\frac{ \pi }{ 6 }}^{\tan^{-1} \frac{ 2\sqrt{3} }{ 3 }}\sqrt{(\sqrt{3}\tan \theta)^{2}+3}\ \sqrt{3} \sec ^{2}\theta d \theta\]Please tell me you know where to go from there.

Answer 11

why is is x = 2 and not x = 3?

Answer 12

Oh sorry, is that a 3? It looked like a 2.

Answer 13

oh okay i understand

Answer 14

OK then the upper limit should be θ = π/3 instead of the inverse tan.

Answer 15

Jenny I'm confused... you didn't include the limits OR the directions in your original post....? :( That makes it quite difficult to help you on these types of problems...

Answer 16

She already admitted the error of her ways, @zepdrix

Answer 17

i originally thought that you could use this with u-sub which appearntly u can't...

Answer 18

and i couldnt find what the heck proprety 8 was until now.

Answer 19

Well you can but then after integration with respect to theta, you would have to do substitution again to rewrite the answer in terms of x. Hence you're only making life unnecessarily difficult.

Answer 20

i ended up using trapezoidal rule.... with 8 subintervals

Answer 21

but lemme try this way.

Answer 22

Alright, that's one way to go. Nothing wrong with that, but you should have mentioned in the beginning that you wanted solve the problem using trapezoidal rule.

Answer 23

my friend suggested it.and i mentioned it somewhere in the other one.

Answer 24

Which way would you like to learn?

Answer 25

Trig substitution is the most efficient method here but what would you like to do? Please pick only one because I really don't have the time to type a lesson for every possible method.

Answer 26

cuz i had already learned trapezoidal rule but not yet learned trig subs so i just stuck to that way. sorry but i'm not gonna learn trig subs till next semester so if it's alright, i'll just let my teacher teach me? cuz i don't want too many things in my head all at once...
i mean i'll look through what u posted, but it'll just be a little hard for me to catch on...

Answer 27

Like I said, I can still teach you trapezoidal rule, if you'd like?

Answer 28

no i had already learned that so that's why i used that

Answer 29

OK, so you got the answer then? You're good?

Answer 30

i got close to the answer because it was an approximation so yeah thanks tho :D

Answer 31

Of course, it is an approximation. OK then I'm glad to hear you got it.