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## Simple Integration Doubt regarding integral of dy

1. The problem statement, all variables and given/known data
What is the result of ∫dx??? is it x or x+C
1) Through indefinite integration, it gives x+C
2) If I take a geometric interpretation, this integral gives me the area under the [f(x) and x] graph where f(x)=1 so by that the integral must be x(is there a C????)

If it were x+C then in the khan academy video
http://www.khanacademy.org/math/calc...tial-equations
He puts
∫dy=y, wouldn't it be ∫dy=y+C

2. Relevant equations

3. The attempt at a solution
Getting confused!

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Let us assume: dy = dx and integrate from 0 to y and a to x respectively. Here the value of a would determine the expression of the function, namely the C in your equation. In the case of x, the a is given as 0. Whereas a is any given number we obtain the result x + C.

Thanks for the reply. I understood what you wrote above and it is quite helpful. But in the case of solving differential equations or in any other application of calculus, will we take it as x or x+C similarly will we take it as y or y+C  (I am self-studying calculus right now so I will be posting many doubts which I have. Help like this will be appreciated) :)

## Simple Integration Doubt regarding integral of dy

If it is in INDEFINITE integration, there will ALWAYS be an arbitrary constant C.
If it is a DEFINITE integration, there is NEVER a C.

If there are other conditions given to you (say the value of the integrate at some point), you might be able to calculate C.

Then, in the Khan academy video, whose link I have given above, shouldn't it be y+C in the first part of the video. If there is, then wouldn't the C's cancel out giving only the variables

Each indefinite integration yields an arbitrary constant C. So ∫dy will give a constant Cy, ∫dx will give another constant Cx1 and ∫x2dx another.   Having said that, Khan academy has absorbed all three of them in one and named it C, which is a valid thing to do as long as they are all additive constants.

In the case of simple differential equation, if we know the initial conditions, we can work out C by substituting values into the expression. It is equivalent to integrate $\int^{y}_{y_{0}}g(y)dy = \int^{x}_{x_{0}}f(x)dx$ where $y_{0} x_{0}$ are the initial conditions.  I must apologise for the ambiguities above. I was trying to express the indefinite integration with definite integration. Both expressions are equivalent in the case

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Quote by sarvesh0303 					 				 				1. The problem statement, all variables and given/known data What is the result of ∫dx??? is it x or x+C I thought about this two ways: 1) Through indefinite integration, it gives x+C 2) If I take a geometric interpretation, this integral gives me the area under the [f(x) and x] graph where f(x)=1 so by that the integral must be x(is there a C????)
The area of what region? You have an upper boundary (y= 1) and lower boundary (y= 0), but have not specified left and right boundaries. If you take some fixed value, $x_0$, as left boundary and the variable x as right boundary, you have a rectangle of height 1 and width $x- x_0$. The integral is $x- x_0$ which is the same as x+ C for C equal to $x_0$.

If it were x+C then in the khan academy video http://www.khanacademy.org/math/calc...tial-equations He puts ∫dy=y, wouldn't it be ∫dy=y+C  2. Relevant equations    3. The attempt at a solution Getting confused!

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Quote by sarvesh0303 					 				 				1. The problem statement, all variables and given/known data What is the result of ∫dx??? is it x or x+C I thought about this two ways: 1) Through indefinite integration, it gives x+C 2) If I take a geometric interpretation, this integral gives me the area under the [f(x) and x] graph where f(x)=1 so by that the integral must be x(is there a C????)  If it were x+C then in the khan academy video http://www.khanacademy.org/math/calc...tial-equations He puts ∫dy=y, wouldn't it be ∫dy=y+C  2. Relevant equations    3. The attempt at a solution Getting confused!
Constants of integration are arbitrary as long as no additional information is fed into the problem. So, if we have and equation of the form dy = dx, we can integrate on both sides to get y + K = x + L, where K and L are two separate constants of integration. We can, of course, re-write this as y = x + C, where C = L - K is also an arbitrary constant. So, for example, y = x, y = x+5, y = x - 3, y = x + 2π, ... all satisfy dy = dx.

RGV

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