Question

Triple Integral at the bottom of the mess of replies. How would I go about solving it?

Answer 1

o_O

Answer 2

\[f(x) = \frac{ 1 }{ 2 }x^4\]
\[f'(x) = \frac{ 1 }{ 2 }4x^3\]
\[f'(x) = 2x^3\]
slope of line tangent to 1/2x^4 at (2,8).
so
now i must turn my sleepy brain on. ..

Answer 3

Function of x?

Answer 4

no i did that wrong oops

Answer 5

@shamil98 We solved a question like this for that unresponsive guy.

Answer 6

oh no i didnt, i should sleep.

Answer 7

yes yes, i know i've done this before just reviewing

Answer 8

now plug in 2 instead of x in that f'(x)

Answer 9

yep.

Answer 10

@ParthKohli I like that dog you used.

Answer 11

@Isaiah.Feynman in the end of ozymandias? that was vince's idea.

Answer 12

Vince's idea was good. @ParthKohli I ORDER YOU TO USE IT NOW!

Answer 13

I was referencing Breaking Bad. Anyway...

Answer 14

oh got it . 16

Answer 15

maybe 1 am is a bad time for this stuff.

Answer 16

@shamil98 LOL

Answer 17

I'm studying statistics its 2 pm

Answer 18

Batman just keeps observing!

Answer 19

I'm on brilliant.org, just solved a level 5 problemmmmm :P

Answer 20

Are you in Pak, @Isaiah.Feynman

Answer 21

get on my level sham

Answer 22

noob <3

Answer 23

Nope. I'm in the northern west.

Answer 24

I did too!

Answer 25

are between curves y = x^2 and y = x^3
from 1 <= x <= 3
\[\int\limits_{1}^{3} x^2 + \int\limits_{1}^{3} x^3\]
x^3/3 )from 1 to 3.. x^4/4 from 1 to 3
3^3/3 - 1^3/3 3^4/4 - 1^4/4
27/3 - 1 /3 81/4 - 1/4
26/3 + 80/4
= 86/3

i did something wrong what did i do wrong.

Answer 26

batman i'm on lvl 3 atm

Answer 27

I was doing algebra, I just finished it, pretty easy tbh.

Answer 28

oh wait i got it

Answer 29

Are you solving for area?

Answer 30

why adding the areas?

Answer 31

because parth i am dumb
\[\int\limits_{1}^{3} x^2 - x^3 ~dx\]

Answer 32

Lmao sham

Answer 33

x^3/3 - x^4/4

(3^3/3 - 3^4/4) - ( 1^3/3 - 1^4/4)
9 - 81/4 - 1/3 + 1/4
9 - 1/3 + 20
29 - 1/3
86/3

Answer 34

wait i got same answer what

Answer 35

oh arithmetic error

Answer 36

-34/3

Answer 37

Darn, it says that is wrong too.

Answer 38

how can area b/w two curves be negative?

Answer 39

You can't have a negative area sham

Answer 40

nvm i made another error it was x^3 - x^2
I was looking at the graphs from x = 0 to x= 1 ....
its 11.3333

Answer 41

Do my failures amuse you all?

Answer 42

I like that you're persistent! :)

Answer 43

How do i do this?..
\[\large \int\limits_{}^{} \int\limits_{}^{} \int\limits_{B}^{} xyz^2 dV\]
where B is the cuboid region bounded by the regions 0 ≤ x ≤ 1, - 1 ≤ y ≤ 2 and
0 ≤ z ≤ 3?

Answer 44

Lol that's on the website right?

Answer 45

yah

Answer 46

level four

Answer 47

I looked at it, and clicked X.

Answer 48

they require work

Answer 49

how do you do it though, the dV is a bit offputting to me >.>

Answer 50

if it was something like dzdydx
i mean that would be pretty simple

Answer 51

you guys have any ideas? i'm going to head to sleep now, got school in a few hours..
but i would appreciate any thoughts on the triple integral, thanks.

Answer 52

=27/4

Answer 53

I think

Answer 54

Was that the full problem?