Question

determine if the improper integral converges or diverges

Answer 1

\[\int\limits_{0}^{4} 1/\sqrt{16-x^2}dx\]

Answer 2

i think you have to use a trig substitution

Answer 3

\[1-\sin ^2\theta=\cos ^2\]

Answer 4

my educated guess would be that it diverges

Answer 5

what are you gonna sub for x ?

Answer 6

well i was thinking about completing the square

Answer 7

how about a trig sub?

Answer 8

\[x=4\sin u\]

Answer 9

thats what i said earlier

Answer 10

you never mentioned the 4 exactly... or du

Answer 11

ok lol my bad

Answer 12

it's cool, so what is du ?

Answer 13

is it -cos?

Answer 14

\[x=4\sin u\implies du=4\cos udu\]

Answer 15

ok

Answer 16

typo*\[x=4\sin u\implies dx=4\cos udu\]so sub that into the integrand and simplify
what do you get under the integral sign?

Answer 17

does the sin/cos become tan?

Answer 18

remember to use the identity you mentioned

Answer 19

\[4\sin \theta^2-1\]

Answer 20

\[{1\over\sqrt{16-(4\sin u)^2}}\cdot4\cos udu=\frac44{\cos u\over\sqrt{1-\sin^2 u}}={\cos u\over\cos u}=1\]

Answer 21

0 to 4?

Answer 22

because we have\[x=4\sin u\]that means our bounds change to be in terms of u\[0=4\sin u\implies u=\sin^{-1}0=0\]\[4=4\sin u\implies u=\sin^{-1}1=\frac\pi2\]so now our integral is from 0 to pi/2

Answer 23

that was weird...

Answer 24

\[\int\limits_{0}^{4}/\sqrt{16-x^2}dx=\int\limits_{0}^{4}(1/\sqrt{16})\sqrt{1-x^2/16}dx\]
\[(1/4)\int\limits_{0}^{4}1/\sqrt{1-x^2/16}dx\]
sub u=x/4 du=dx/4 dx=4du
\[\int\limits_{0}^{4}1/\sqrt{1-u^2}du=\sin^{-1}u|_{0}^{4}\]
\[\sin^{-1}(x/4)|_{0}^{4}=\sin^{-1}(1)-\sin^{-1}(0)\]

Answer 25

yeppers

Answer 26

typo"
\[\int\limits_{0}^{4}(1/\sqrt{16})/\sqrt{1-x^2/16}dx\]

Answer 27

or with what I gave you you could just say that the integral is\[\int_0^{\pi/2}du\]

Answer 28

ok i understood when you did substitution and got cosu/cosu=1

Answer 29

but i know that the bound is u but i dont get what you did in that part

Answer 30

our original bounds are \(x=0\) and \(x=4\) but we made the substitution\[x=4\sin u\]so now we want to know what u is for each x bound
first let's find what u is when x=0...

Answer 31

\[x=4\sin u\]so\[0=4\sin u\]solve this for u\[\sin u=0\implies u=\sin^{-1}0=0\]so that is our first bound; i.e. when x=0, u=0 also
now for the other bound, x=4...

Answer 32

\[x=4\sin u\]subbing in the bound x=4 we get\[4=4\sin u\]solving for u we get\[1=\sin u\]\[u=\sin^{-1}(1)=\frac\pi2\]which is our new upper bound

make sense?

Answer 33

ok yes im still digesting this last part were u=sin^-1(1)=pi/2 but yes i got the first part

Answer 34

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
gotta remember your special values for that one
check out the unit circle part of this link for a refresher

Answer 35

oooh yeah it clicked

Answer 36

so now you can see how we reduced this ugly-loking thing to\[\int_0^{\pi/2}du\]and so it converges to...?

Answer 37

what do i replace du with?

Answer 38

nothing, just integrate

Answer 39

...and evaluate

Answer 40

u+c?

Answer 41

the constant only shows up in indefinite integrals
it goes away when we evaluate
what does it evaluate to?

Answer 42

pi/2?

Answer 43

right :)
questions? comments?

Answer 44

wow...lol sorry for the slow mode. but no none :) so it converges

Answer 45

yes, to exactly \(\large \frac\pi2\)
amazing ain't it? :D

Answer 46

yes quite strange how everything just simplified

Answer 47

that's what you call "elegance" in math
it makes it purty...

Answer 48

lol thanks again! lots of help! :)

Answer 49

very welcome!