Question

improper integral...

Answer 1

\[\int\limits_{2}^{4} \frac{ dx }{ \sqrt{4x-x ^{2}} } \]

Answer 2

how do i find if the integral diverges or converges and its value?

Answer 3

use a trig sub

Answer 4

you will need to complete the square first of course

Answer 5

how do i complete the square?

Answer 6

\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

Answer 7

im confused. do i have to square the square root of 4x-x^2?

Answer 8

how do you mean?

Answer 9

\[-x^2+4x=-(x^2-4x)\]
\[=-(x^2-4x+?)+?\]

You must complete the square to determine what trig sub to use...
Do not square the square root of 4x-x^2...

Answer 10

have you figure out what number to replace the ? with to complete the square

Answer 11

see above formula if you are still having trouble figuring out the value for ?

Answer 12

hmm

Answer 13

k/2 ?

Answer 14

you do realize we have 4's instead of k's right? :p

Answer 15

ohh. 2?

Answer 16

also it should be (k/2)^2
where k is -4

Answer 17

ohh 4?

Answer 18

\[-x^2+4x=-(x^2-4x) \\ =-(x^2-4x+(\frac{-4}{2})^2)+(\frac{-4}{2})^2 \\ \]

Answer 19

now you are above to complete the square for the x terms

Answer 20

\[=-(x-2)^2+(\frac{-4}{2})^2\]
\[=-(x-2)^2+4\]

Answer 21

so your integral should look like
\[\int_2^4 \frac{dx}{\sqrt{4-(x-2)^2}}\]

now do a trig sub

Answer 22

\[\sqrt{a^2-x^2} ,x = a \sin \theta \]
\[\sqrt{2^{2}-(x-2)^{2}} , x=2\sin \theta \]
is that right? :/

Answer 23

not exactly
you have x-2 inside the square not just x

Answer 24

x-2=2 sin(theta)

Answer 25

ok. x = 2 sin(theta) +2
then do i take the derivativeof that?

Answer 26

yes

Answer 27

=2cos(theta)

Answer 28

do you mean dx=2 cos(theta) d theta ?

Answer 29

ahh yes

Answer 30

and then i replace them?

Answer 31

you are trying to write your integral in terms of your substitution
so yes
don't forget to change the limits accordingly

Answer 32

do i plug in x=2sin(theta)+2 into sqrt(4-(x-2)^2 ?

Answer 33

you see the x-2 inside the square?

Answer 34

we called that 2 sin(theta) earlier ....

Answer 35

but yes you can replace x with 2sin(theta)+2

Answer 36

\[\int\limits_{?}^{?}\frac{ 2\cos \theta }{ \sqrt{4-(2\sin \theta)^{2}}} d \theta \]
now to find the limits...

Answer 37

you had x=2 to x=4
and recall we made the sub x-2=2 sin(theta)

solve for theta for both x's
I would use the first positive theta that satisfies the equation

Answer 38

im confused :(

Answer 39

is it 0 and 1.57?

Answer 40

I think you mean from theta=0 to theta=pi/2

Answer 41

if so yes

Answer 42

im not sure how to use the identity 1-sin^2 theta = cos^2 theta

Answer 43

the identity still holds if you multiply both sides by 4
\[4-4 \sin^2(\theta)=4 \cos^2(\theta)\]

Answer 44

thank you. i got it. pi/2

Answer 45

great job