Question

Integration. Easy one.

Answer 1

\[\int\limits_{-2}^{2}4(4-X^2)^1/2 dx\]

Answer 2

The answer is 8pi

Answer 3

is this not easier with substitution?

Answer 4

do u need help with steps

Answer 5

\(\color{#000000 }{ \displaystyle \int_{-2}^{2}~4(4-x^2)^{1/2}dx }\) ?

Answer 6

like that

Answer 7

let x=2sin(theta)

Answer 8

Yes, correct.

Answer 9

yes @SolomonZelman

Answer 10

http://www.wolframalpha.com/input/?i=integrate+4%284-x%5E2%29%5E1%2F2+from+-2+to+2+

Answer 11

and then it turns out to be \[16 \int\limits_{-2}^{2} \cos^2(\theta) d(\theta)\]

Answer 12

This is not giving me 8pi form wolfram so i guess I make mistakes along the way.

Answer 13

yes @ayeshaafzal221 yes please.

Answer 14

i will do step by step

Answer 15

\[\sqrt{4-x^2} \] sub x=2sin(u) and du=2cos(u) du

Answer 16

yes I did all that @ayeshaafzal221 .

Answer 17

then \[\sqrt{4-x^2 }=\sqrt{4-4\sin^2 (u)}=2\sqrt{\cos^2 (u)}\]

Answer 18

did u converted the bounds ?

Answer 19

oh NOOOOOOOOOOO I DID NOT

Answer 20

I kept using 2 and -2

Answer 21

\[-\frac{ \pi }{ 2 }<u<\frac{ \pi }{ 2 }\]

Answer 22

How did you get that?

Answer 23

i ll show u

Answer 24

with inverse \[u=\sin^{-1}(\frac{ x }{ 2 })\]

Answer 25

Still did'nt get it?

Answer 26

this will give lower bound \[u=\sin^{-1} (-\frac{ 2 }{ 2 })=-\frac{ \pi }{ 2 }\]

Answer 27

and upper bound \[u=\sin^{-1} (\frac{ 2 }{ 2 })=\frac{ \pi }{ 2 }\]

Answer 28

did u get it ?

Answer 29

Oh yea I do, sub x into x=2sin(theta).

Answer 30

and -2 for the lower bound.

Answer 31

thats right :) well done

Answer 32

Let me try one more time. :)

Answer 33

\(\color{#000000 }{ \displaystyle \int_{-2}^{2}~4(4-x^2)^{1/2}dx }\)

\(\color{#000000 }{ \displaystyle x=2\sin \theta }\)
\(\color{#000000 }{ \displaystyle dx=2\cos \theta~d \theta }\)

\(\color{#000000 }{ \displaystyle \int_{-2}^{2}~8(4-4\sin^2\theta)^{1/2}\cos \theta~d \theta }\)

\(\color{#000000 }{ \displaystyle\int_{-2}^{2}~16\cos^2\theta ~d \theta }\)

You know;
\(\color{#000000 }{ \displaystyle \cos (2z)=\cos^2z - \sin^2 z }\)
\(\color{#000000 }{ \displaystyle \cos (2z)=\cos^2z -(1-\sin^2 z) }\)
\(\color{#000000 }{ \displaystyle \cos (2z)=2\cos^2z -1 }\)
\(\color{#000000 }{ \displaystyle \cos (2z)+1=2\cos^2z }\)
\(\color{#000000 }{ \displaystyle (1/2)\cos (2z)+ (1/2)=\cos^2z }\)

\(\color{#000000 }{ \displaystyle\int_{-2}^{2}~8\cos(2\theta)+8 ~d \theta }\)

\(\color{#000000 }{ \displaystyle \left.4\cos(2\theta)+8\theta \color{white}{\LARGE|}\right|_{-2}^{2} }\)

\(\color{#000000 }{ \displaystyle \left.8\cos\theta\sin \theta +8\theta \color{white}{\LARGE|}\right|_{-2}^{2} }\)

the limits belong to x, not \(\theta\).
\(\color{#000000 }{ \displaystyle \sin^{-1}(x/2)= \theta }\)

\(\color{#000000 }{ \displaystyle \left.8\cos(\sin^{-1}(x/2))\sin (\sin^{-1}(x/2)) +8\sin^{-1}(x/2) \color{white}{\LARGE|}\right|_{-2}^{2} }\)

\(\color{#000000 }{ \displaystyle \left.(4x)\cos(\sin^{-1}(x/2)) +8\sin^{-1}(x/2) \color{white}{\LARGE|}\right|_{-2}^{2} }\)

\(\color{#000000 }{ \displaystyle \left.(4x)\sqrt{1-\sin(\arcsin^{2}(x/2))} +8\sin^{-1}(x/2) \color{white}{\LARGE|}\right|_{-2}^{2} }\)

\(\color{#000000 }{ \displaystyle \left.(4x)\sqrt{1-x^2/4} +8\sin^{-1}(x/2) \color{white}{\LARGE|}\right|_{-2}^{2} }\)

\(\color{#000000 }{ \displaystyle \left.(2x)\sqrt{4-x^2} +8\sin^{-1}(x/2) \color{white}{\LARGE|}\right|_{-2}^{2} }\)

Answer 34

that is the result of integration, but you can do it geometrically

Answer 35

@SolomonZelman why didnt u convert the bounds ?

Answer 36

Nope I did not

Answer 37

I didn't change the limits of integration

Answer 38

to verify the result of indefinite integral that wolfram has

Answer 39

Anyway... back to geom....

Answer 40

anyways @khally92 just tag me if u need further help :)

Answer 41

oh sure I will one more thing @ayeshaafzal221 . infact very soon. thanks. you want the medal. I dont think @SolomonZelman really cares about medal.

Answer 42

\(\color{#000000 }{ \displaystyle y=\sqrt{4-x^2} }\)

That is a half-circle,
\(\color{#000000 }{ \displaystyle y^2=4-x^2;\quad\quad \left\{y\ge0\right\} }\)
\(\color{#000000 }{ \displaystyle y^2+x^2=2^2;\quad\quad \left\{y\ge0\right\} }\)

So, therefore,
\(\color{#000000 }{ \displaystyle \int_{x=-2}^{x=2}\sqrt{4-x^2}dx=\frac{1}{2}\cdot \pi(2^2)=2\pi }\)

And thus,

\(\color{#000000 }{ \displaystyle 4\int_{x=-2}^{x=2}\sqrt{4-x^2}dx=8\pi }\)

Answer 43

Area of a circle is pi r^2

Answer 44

But, area of half-circle ... ?

Answer 45

And our r here is 2

Answer 46

yeah I know. so pi r^2 gives you 4pi

Answer 47

and divide by 2 gives you 2pi??

Answer 48

yes; do you see why, or is that a question to me to address?

Answer 49

you see what I did in this geometric approach, or not?

Answer 50

oh I forgot about the 4 outside.

Answer 51

@ayeshaafzal221 is (pi/2) and -(pi/2) in rad or deg. I keep getting confuse.

Answer 52

Thanks @SolomonZelman really appreciate. I am doing calculus 3 and it seems like I have to brush up on my integrals

Answer 53

@khally92 in rad

Answer 54

This is a pretty hard integral to take; and if you didn't remember to substitute the original variable back or something of this sort then it is not a catastrophy.

The only reason I did the integral way is because you asked, but the gometry came first to my mind, and this is what you should have also done. You should have noticed the half-circle...

Answer 55

I am not a professor, mathematician, or anything of the sort, but I would advise to go over circles/half-circles, and to do some integrals occasionally too.

Answer 56

gracias @SolomonZelman I will. Thanks.

Answer 57

yw

Answer 58

@ayeshaafzal221 I will use the geometry its much easier but i do still want to get this, but its not giving me 8pi. see what I have \[16\int\limits_{-\pi/2}^{\pi/2} \cos^2\theta d \theta = 16 [1/2\sin \theta \cos \theta + \theta]\]

Answer 59

the 1/2 also multiplies theta

Answer 60

oh forget it thanks.

Answer 61

got it.

Answer 62

ur last answer should be \[\int\limits_{-\frac{ \pi }{ 2 }}^{\frac{ \pi }{ 2 }} 8u\]

Answer 63

ok good :)