Question

Find the area:

Answer 1

\[\large \int\limits_{0}^{\pi} sinx~dx\]

Answer 2

Bored?

Answer 3

And reviewing ;P

Answer 4

Antiderivative of sin(x) is...?

Answer 5

Lol, okay then. Integral sinx? We remember that?

Answer 6

Uh, -cosx?

Answer 7

damnit this problem is easy D:

Answer 8

Well, if you're that bored, you might as well use
\[\Large e^{2\pi^3}-\cos(x)\]

Answer 9

Lol.

Answer 10

\[\int\limits_{}^{}e ^{x^{2}}dx\]

Answer 11

Well.... there's nothing we can do about that ^
Unless you're suggesting series...

Answer 12

Psymon, what's a double integral?

Answer 13

This:
\[\Large \iint\limits_R f(x,y)dA\]? :D

Answer 14

Lol, actually, there is something you can do about the one I posted. Thats why I specifically posted that one.

Answer 15

None that I know of. :/

Answer 16

Its actually a special function. Its called an error function, lol. And then theres also a complementary error function.

Answer 17

Theres all these really weird special functions Im being shown in class :/

Answer 18

\[F(\pi) - F(0)\]
\[-\cos(\pi) - [-\cos(0)]\]
@Luigi0210

Answer 19

@Psymon You forgot something :3
\[\Large \color{green}{e^{2\pi^3}}-\cos(\pi) - [\color{green}{e^{2\pi^3}}-\cos(0]\]

Answer 20

Not in reference to yours D: @terenzreignz

Answer 21

Hey Psymon is this right?
\[\int\limits_{0}^{1}(\int\limits_{0}^{2}xy^2~dx)dy\]
\[\int\limits_{0}^{1}(2y^2)dy=\frac{2}{3}\]

Answer 22

Yep. I think so at first glance.

Answer 23

Didnt do any paper, though.

Answer 24

I barely know how to do single integrals o_O

Answer 25

Double and triple integrals are easy. Same with partial derivatives. Its only when you have to start doing conversions an dsubstitutions with the integrals that things get interesting.

Answer 26

Well for you, I have very little experience with integrals

Answer 27

You did it right, though O.o

Answer 28

\[\int\limits_{0}^{1}\int\limits_{0}^{2}(xy^2)~dxdy=\int\limits_0^1x~dx\int\limits_0^2y^2dy\]

Answer 29

Is it really back integral first? O.o

Answer 30

Doesn't matter since the x and y expressions are separable from each other.

Answer 31

Okay, yeah, i havent ever worked with doubles and triples, I just know the concept xD So you show me something and its likely to benew, haha.

Answer 32

Well, there's something wrong
\[\Large \frac12\cdot \frac{8}3\]

Answer 33

\[\Large = \frac43 \ne \frac23\]

Answer 34

Eh O.o

Answer 35

Well, I know what luigi did to get his answer, so what did you do.

Answer 36

And what do you get when you integrate y^2? -.-

Answer 37

Sorry I haven't slept xD

Answer 38

Unless... I got it reversed

Answer 39

Okay, so seriously, why is it multiplication instead of what im thinking is going on?
\[\int\limits_{0}^{1}\int\limits_{0}^{2}xy^{2}dxdy\]
\[\frac{ x^{2} y^2}{ 2 }\]
F(2) - F(0) = 2y^2

\[\int\limits_{0}^{1}2y^{2}dy \]
\[\frac{ 2y^{3} }{ 3 } \]
F(1) - F(0) = 2/3

What am I not getting?

Answer 40

My mistake :D

Answer 41

Sorry, must have not integrated with the correct limits... must have switched them

Answer 42

hehe... whoops

Answer 43

Lol, no biggie xD

Answer 44

._.

Answer 45

Luigi be correcti :3

Answer 46

I'm going back to algebra >.>

Answer 47

Hey, you got it, no need to xD Why the double integral question, though?

Answer 48

I'm logging out... I've caused enough trouble for today :3
Catch you guys later ^_^
------------------------------------
TJ out

Answer 49

I was curious? Idk -_-
See yall TJ

Now.. how do you graph this? >.<

Answer 50

xy^2?

Answer 51

Yea, whatever that means :/

Answer 52

Im not sure really. It doesnt fit any of the forms I would know how to graph, haha x_x

Answer 53

I mean, I can kinda guess.

Answer 54

\[z=xy^2\]
\[0 \le x \le2 \]
\[0\le x \le 1\]
My calculuator gave me this >.>

Answer 55

*\[0 \le y \le 1\]

Answer 56

I couldnt tell ya what my graphing calculator got. TOo hard to perfectly tell, lol.

Answer 57

I need to go back to watching videos >.<
Thanks Psymon :)

Answer 58

Alright, laterz xD