Simple Integration Doubt regarding integral of dy

[sarvesh0303]

Simple Integration Doubt regarding integrals

Homework Statement

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/calculus/integral-calculus/v/simple-differential-equations
He puts
∫dy=y, wouldn't it be ∫dy=y+C

Homework Equations

The Attempt at a Solution

Getting confused!

 

[raopeng]

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.

 

[sarvesh0303]

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) :)

 

[Sourabh N]

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.

 

[sarvesh0303]

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

 

[Sourabh N]

Each indefinite integration yields an arbitrary constant C. So ∫dy will give a constant C_y, ∫dx will give another constant C_x1 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.

 

[raopeng]

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

 

[HallsofIvy]

Homework Statement

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/calculus/integral-calculus/v/simple-differential-equations
He puts
∫dy=y, wouldn't it be ∫dy=y+C

Homework Equations

The Attempt at a Solution

Getting confused!

 

[Ray Vickson]

Homework Statement

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/calculus/integral-calculus/v/simple-differential-equations
He puts
∫dy=y, wouldn't it be ∫dy=y+C

Homework Equations

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