Question

Evaluate the integral: Sqrt[4-x^2], n= 0 to 2

Answer 1

That's beatiful :-)

Answer 2

use a trig. substitution. theta=2sintheta

Answer 3

@bmp Lmao, what?

Answer 4

The answer to it... Anyway, use x = 2sin(u) and dx = 2cos(u)du. The square root becomes: \[ \sqrt{4-x^2} = \sqrt{4-4\sin^2{u}} = 2\cos(u) \]

Answer 5

Not necessary. Think of it as an area of a quarter circle with radius 2.

Answer 6

Using polar coordinates becomes too easy :-). But, yes, I agree with @blockcolder . That's a lot easier.

Answer 7

\[\sqrt{a ^{2}+x ^{2}} use x=asine(\theta)\]
\[\sqrt{a ^{2}-x ^{2}} use x=atan(\theta)\]
\[\sqrt{x ^{2}-a ^{2}} use x=asec(\theta)\]

Answer 8

Doesn't make the answer less pretty, tho. :-)

Answer 9

im sure the teacher want to see trig subs, this is for his finals lol =) just saying

Answer 10

Well then: Hint: Write \[ \int 4\cos^2{(u)}du = 4 \int (\frac{1}{2} \cos(2u) + \frac{1}{2})du\]and integrate term by term.

Answer 11

TYPO---- last x=asec(theta),, its + not -..... !!!!!

Answer 12

How on earth do u find this stuff fun?!!? @bmp I'll stick to what you wrtoe before.

Answer 13

omg my signs is asll wrong lol ignroe it. ill rewrite it

Answer 14

You dont have to. I got it.

Answer 15

ok

Answer 16

Calculus is, very likely, the most fun subject as an undergrad for me.

Answer 17

Lol, genius. I just want to finish this course and change my major

Answer 18

No, not at all. I suck at Linear Algebra, for instance. I barely passed it. Change it to Mathematics :-) Just kidding.

Answer 19

lmao! Yeaaah, that was definitely good joke.

Answer 20

even my grammar is messed up now

Answer 21

My native language is messed up too. I think it's a consequence of doing something related to an exact sciences.