Question

So I just heard my advisor lecture his students that performing a double integral on an expression is not the same thing as integrating that expression twice (all with respect to a single, independent variable).

The context of the problem was the solution of an autonomous, 2nd order ODE:
\[\frac{d^2y}{dx^2}=f(x).\]Apparently,
\[\int \int \frac{d^2y}{dx^2}=\int \int f(x)\] is not the same thing as
\[\int [ \int \frac{d^2y}{dx^2} ]=\int [ \int f(x) ]\]Why did he say that?

Answer 1

i don't know i always work inside to outside
i don't see how it isn't the same (yuck double negative)
i see it as the same

Answer 2

He further added, although rather ambiguously, that a second integral had more to do with spatial concepts.

I didn't want to question him for fear that my students would lose trust in my reviews during TA sessions, had I been wrong in some remote, arcane sense.

Perhaps they're not the same thing when viewed from a more convoluted position?

Answer 3

so he didn't have any limits?

Answer 4

i remember you can't switch the outside integral with the inside integral sometimes

Answer 5

There were no limits; the problem was stated just as I wrote it.

Answer 6

imran any thoughts?

Answer 7

no , my brain is melting from physics

Answer 8

physics>math ?

Answer 9

If you know quantum physics, talk to me. xd

Answer 10

no, just classical

Answer 11

ok i'm going to see if i can find anything on your double integral question

Answer 12

I think we can switch order if limits are constant, or something

Answer 13

I totally agree with you that they're not the same thing when limits are incorporated. Perhaps he just said that to discourage students from thinking about it that way?

Answer 14

f(x)=5x
\[\int\limits_{}^{}\int\limits_{}^{}(5x) dx dx=\int\limits_{}^{}(\frac{5x^2}{2}+C)dx=\frac{5x^3}{2(3)}+Cx+D=\frac{5x^3}{6}+Cx+D\]
is this incorrect?

Answer 15

It's correct.

Answer 16

but i did this
\[\int\limits_{}^{} [\int\limits_{}^{}5x dx] dx\]

Answer 17

i performed the inside integral first
then the outside

Answer 18

You know what, I think he may have just messed up because, as far as I know, that's how you solve a double integral.

Answer 19

what was he integrating with respect to...did he leave that off like you wrote?

Answer 20

He stated we were talking about one dependent and one independent variable. So the integral (in this case) was (implicitly) with respect to x.

Answer 21

I'll ask him about this and see what reason he gives me.

Answer 22

I only see a difference when there are limits dxdy vs dA

Answer 23

\[\int\limits_{}^{}\int\limits_{}^{}(5x)dxdy=\int\limits_{}^{}(\frac{5x^2}{2}+C)dy=\frac{5x^2y}{2}+Cy+D\]
\[\neq\]
\[\int\limits_{}^{}\int\limits_{}^{}(5x)dydx=\int\limits_{}^{}(5xy+C) dx=\frac{5x^2y}{2}+Cx+D\]
maybe he left the limits off because he was trying to be general

Answer 24

not the limits

Answer 25

but what we are integrating with respect too

Answer 26

ok i have to figure out how to program a line in c++

Answer 27

eventhough imran wants me to use cg

Answer 28

we need to be a little careful when we do things like...

\["\int\limits_{}^{}\int\limits_{}^{}(5x)dxdy=\int\limits_{}^{}(\frac{5x^2}{2}+C)dy=\frac{5x^2y}{2}+Cy+D"\]

if we have a two variable function then

\[\int\int 5x \,\,dxdy=\int\left(\frac{5}{2}x^2+g(y)\right)dy\]

since \[\frac{\partial}{\partial x}\left(\frac{5}{2}x^2+g(y)\right)=5x^2\]

Answer 29

you are right

Answer 30

gj zarkon!
i would give you another medal if i could

Answer 31

As far as I know, he was wrong. (I don't see any reason the brackets modify the scenario.) I also know that this input is not exactly relevant 10 months later.

Answer 32

nobody noticed my typo ;)

\[\frac{\partial}{\partial x}\left(\frac{5}{2}x^2+g(y)\right)=5x\]