Question

∫ sqrt(4-x^2) dx on the closed interval [-2,2].
I know how to do this with graphing it (it's a semicircle, so you would use the equation for a circle, get radius use area for circle formula and so on..etc ).. but I was wondering how to do this without graphing it?..

Answer 1

The answer is C

Answer 2

This isn't a multiple choice question, nor is it for homework. -_-

Answer 3

I am simply wondering how it is done.

Answer 4

XP i know

Answer 5

If you are unable to provide any insight please refrain from posting further irrelevant comments.

Answer 6

well then,

Answer 7

one way is to substitute
x=2sin(u)
dx=2cos(u)du
integrate(sqrt(4-x^2))
=integrate(4cos(u)^2)
=integrate(2+2cos(2u)) from u=-pi/2 to u=pi/2

Answer 8

Agreed.

Answer 9

xD lol

Answer 10

i'm a bit confused, probably cause latex is down and its harder to understand what is being said..

Answer 11

∫ sqrt(4-x^2)
why does it turn to
∫ (4cos(u)^2)
the sqrt symbol just disappears ?
I haven't done much substitution problems yet so i'm not that familiar with how it works yet.

Answer 12

integrate(sqrt(4-x^2)dx)
=integrate(sqrt(4-(2sin(u))^2)dx)
=integrate(sqrt(4-4sin(u)^2)dx)
=integrate(2sqrt(1-sin(u)^2)dx)
=integrate(2sqrt(cos(u)^2)dx)
=integrate(2cos(u)dx)
=integrate(2cos(u)*2cos(u)du)

Answer 13

the trick is substituting x = 2 sin u
dx = 2 cos u du

Answer 14

Shamil: I admire your courage and curiosity in trying to learn another way to do this problem. However, when you go beyond what you're familiar with, you're bound to run into things you don't understand yet (which has happened in this case). Hope you'll continue taking risks, but be aware that there are risks involved.

Answer 15

@shamil,
an identity in trigono, that
1-sin^2 θ = cos^2 θ

so, the integral becomes
int sqrt (4cos^2 θ ) * 2 cosθ dθ
obvious, sqrt (4cos^2 θ ) = 2cosθ, right ?

Answer 16

ah, that makes more sense thanks rad.

mathmale, thanks for the input, I should probably get more familiar with trig sub first then.

Answer 17

trig sub is always the pain the neck :DD