ques

- anonymous

ques

- Stacey Warren - Expert brainly.com

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- chestercat

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- rainbow_rocks03

What is your question?

- anonymous

How to solve an integral of the type
\[\int\limits (f(x,y)dx+g(x,y)dy)\]

- hartnn

distribute the integral
\(\int f + \int g\)
integrate f w.r.t x
treating y as a constant!
integrate g w.r.t y,
treating x as constant!

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## More answers

- anonymous

I tried that but it turned out to be wrong answer

- hartnn

initial conditions are given?
or what other conditions are given?
when you integrate f w.r.t x,
you get some function in x and y
+ some constant
the thing to note here is.,
"the constant" here is actually the function of y !

- anonymous

\[f(x,y)=y^2\]\[g(x,y)=2xy\]\[\int\limits f(x,y)dx+\int\limits g(x,y)dy=\int\limits y^2dx+\int\limits 2xydy=xy^2+xy^2=2xy^2+C\]
But answer is \[xy^2+C\]

- hartnn

that goes like this:
\(\int fdx = xy^2 +f(y)\)
\(\int gdy = xy^2 +f(x)\)
comparing those 2, we can conclude there are no extra functions(terms) which only have y or x
so f(y) = 0, f(x) = 0

- anonymous

\[\vec F=(x^2-y^2+x)\hat i-(2xy+y)\hat j\]
\[d \vec r=dx \hat i+dy \hat j+dz \hat k\]
\[\phi=\int\limits \vec F.d \vec r=\int\limits [(x^2-y^2+x)dx-(2xy+y)dy]+C\]
Instead of -xy^2 I get -2xy^2

- ikram002p

the answer u have from ur book or some where else ??

- ikram002p

i still think in both questions ur answer is correct hmmm how it could be something else :-\

- anonymous

Answer is
\[\frac{x^3}{3}+\frac{x^2}{2}-\frac{y^2}{2}-xy^2+C\]
and I'm getting
\[\frac{x^3}{3}+\frac{x^2}{2}-\frac{y^2}{2}-2xy^2+C\]

- hartnn

x^3/3 -xy^2 +x^2/2 + f(y) ----(A)
-xy^2 +y^2/2 +f(x) ----(B)
comparing the above 2,
we get
f(y) = y^2/2
and
f(x) = x^3/3 +x^2/2
now substitute that in (A) or (B)
to get your answer

- hartnn

-y^2/2 *

- anonymous

Why are we comparing them?Is it necessary that they are equal?

- hartnn

thats how I was taught, didn't raise this question then...
maybe ikram can give some more insight into this...

- anonymous

From what I understand they've done something like
Let
\[h \equiv h(x,y)\] be defined such that
\[f(x,y)=\frac{\partial h}{\partial x}\]
and
\[g(x,y)=\frac{\partial h}{\partial y}\]
so that
\[f(x,y)dx+g(x,y)dy=\frac{\partial h}{\partial x}dx+\frac{\partial h}{\partial y}dy=dh\]
\[\implies \int\limits [f(x,y)dx+g(x,y)dy]=\int\limits dh=h+C\]
But why and even then how would we find our h, it's not always obvious in all cases

- anonymous

One can see at a glance that
\[-\int\limits (y^2dx+2xy dy)=-\int\limits d(xy^2)=-xy^2+C\]

- anonymous

I'll leave this question open, maybe
@IrishBoy123 would be able to help out later,
the real deal is how we will find our h(x,y) and why can't we just simply integrate while keeping one variable constant

- anonymous

One way I thought if since x and y are independent of each other we have
\[\frac{dx}{dy}=\frac{dy}{dx}=0\]
Dividing dh by dx we have
\[\therefore \frac{dh}{dx}=\frac{\partial h}{\partial x} \implies dh=\frac{\partial h}{\partial x}dx=f(x,y)dx\]
Similarly for dy we have
\[dh=g(x,y)dy\]
Adding these two we get
\[2.dh=f(x,y)dx+g(x,y)dy\]
\[2h=\int\limits f(x,y) dx + \int\limits g(x,y) dy\]
Here we mean integration while keeping other variable constant so we will in fact write
\[2h=\int\limits f(x,y) \partial x+\int\limits g(x,y) \partial y\]
\[\implies h=\frac{\int\limits f(x,y) \partial x+\int\limits g(x,y) \partial y}{2}+C\]
Where C is some constant independent of x or y, I'm not sure if it's correct but that's a way to think about it

- anonymous

calculus is weird sometimes lol

- IrishBoy123

these are *line integrals* so you can either solve them as DE's a shown above or you integrate *over a path*. ie if you are to integrate, be careful. as you are travelling over a path, there is actually only one independent variable in line integrals, hence why people often parameterise line integrals into an expression in t only. that's often OTT but it at least is a mechanism that gets answers.
doing that here for the first one;
\(\int y^2 dx+\int 2xy \ dy\), we choose the following path in 2 steps [because we can].
|dw:1440446154961:dw|
for Part 1 of path
y = dy = 0 so the integral is \(\int_{0}^{x} 0^2 \ dx = 0\)
for Part 2
dx = 0, x = const so the integral is \(\int_{0}^{y} 2xy \ dy = xy^2\)
adding gives \(\Phi = xy^2 = const. \)
for the second one:
\(\int x^2−y^2+x) \ dx + \int −(2xy+y) \ dy\)
we chose the same (x,y) destination but go a different way just to illustrate:
|dw:1440446941344:dw|
this time
part 1, dx = x = 0, integral is \(\int_{0}^{y} -y \ dy = -\frac{y^2}{2}\)
part 2, dy = 0, y = const, integral is \(\int_{0}^{x} x^2 - y^2 + x \ dx = \frac{x^3}{3} - xy^2 + \frac{x^2}{2} \)
adding gives \(\Phi = \frac{x^3}{3} - xy^2 + \frac{x^2}{2} -\frac{y^2}{2} = const.\)
you can choose your path because these are conservative vector fields, which is why the also give rise to exact differential equations. furthermore, and this is useful, you can always check you answer by looking at \(\nabla \Phi = \vec F\). ie the gradient of the supposed solution should correspond to the integrands of line integral you started with.
if this is news to you, then honestly, forget about it and solve the way @hartnn was doing above.

- anonymous

thanks I understand it now
first we will check if
\[\vec \nabla \times \vec F=\vec 0\]
If it is true then F is a conservative vector field, so to find the scalar potential phi, such that
\[\vec F=\vec \nabla\phi\]
We can find
\[\phi=\int\limits \vec F.d\vec r\]
along any path in the vector field. Taking a path along the axes or parallel to the axes is most suitable as a variable becomes 0 or constant.

- IrishBoy123

@Nishant_Garg yeah, good summary!

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