3 different solutions to the same integral

[Karol]

Homework Statement

Homework Equations

$$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$

The Attempt at a Solution

If the interpretation of an integral is the derivative at that point, x₁ for example, then these can't all be solutions since the tangent, the derivatives, at x₁ are different.
The constant C can't compensate for that since it shifts the graph up or down, not to the sides, as the solutions $~\sin^2x+C_1$ and $~\cos^2x+C_2$ are, but the solution $-\frac{1}{2}\cos(2x)+C_3$ is neither that nor that.

 

[PeroK]

If the interpretation of an integral is the derivative at that point ...

An integral is not a derivative. It's an anti-derivative. If you differentiate two functions that differ by a constant, you get the same derivative

 

[Karol]

An integral is not a derivative. It's an anti-derivative. If you differentiate two functions that differ by a constant, you get the same derivative

My mistake, of course an integral is an anti-derivative. if i differentiate these 3 functions i indeed get $~2\sin x\cos x~$, how come? indeed the slopes to the graphs of these functions differ for the same x. so my question is can 3 different functions have the same derivative?

 

[PeroK]

My mistake, of course an integral is an anti-derivative. if i differentiate these 3 functions i indeed get $~2\sin x\cos x~$, how come? indeed the slopes to the graphs of these functions differ for the same x. so my question is can 3 different functions have the same derivative?

$f(x) = x+1$
$g(x) = x+2$
$h(x) = x + 3$

Three different functions. Same derivative.

 

[Mark44]

My mistake, of course an integral is an anti-derivative. if i differentiate these 3 functions i indeed get $~2\sin x\cos x~$, how come? indeed the slopes to the graphs of these functions differ for the same x. so my question is can 3 different functions have the same derivative?

In fact, an infinite number of functions can have the same derivative, provided that all of the functions are vertical translations of one another. For example, y = x², y = x² + 1, y = x² + 2, etc. all have exactly the same derivative.

 

[PeroK]

And, of course:

$f(x) = \sin^2(x)$
$g(x) = -\cos^2(x)$
$h(x) = -\frac12 \cos(2x)$

 

[Karol]

Indeed:

Strange, since these functions without the cubes look different

 

[Ray Vickson]

Homework Statement

View attachment 114571 View attachment 114572

Homework Equations

$$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$

The Attempt at a Solution

If the interpretation of an integral is the derivative at that point, x₁ for example, then these can't all be solutions since the tangent, the derivatives, at x₁ are different.
The constant C can't compensate for that since it shifts the graph up or down, not to the sides, as the solutions $~\sin^2x+C_1$ and $~\cos^2x+C_2$ are, but the solution $-\frac{1}{2}\cos(2x)+C_3$ is neither that nor that.

You seem to have forgotten the basic trigonometric identities.

 

[Karol]

You seem to have forgotten the basic trigonometric identities.

I know: $~\sin^2 x+\cos^2 x=1~\rightarrow~\sin^2x=1-\cos^2 x$
And not just $~-\cos^2 x~$.
$$\cos(2x)=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x$$

 

[Ray Vickson]

I know: $~\sin^2 x+\cos^2 x=1~\rightarrow~\sin^2x=1-\cos^2 x$
And not just $~-\cos^2 x~$.
$$\cos(2x)=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x$$

Now you have forgotten the constants of integration. We have $\cos^2 x + C_1 = -\sin^2 x + C_2$ for some constants of integration $C_1,C_2$. There is no contradiction. You will notice in the book extract you included that the book used three separate constants of integration.

 

[Karol]

But:
$$-\frac{1}{2}\cos(2x)\neq -2\sin^2 x+C$$

 

[PeroK]

But:
$$-\frac{1}{2}\cos(2x)\neq -2\sin^2 x+C$$

See post #6 for the correct relationship.

 

[Karol]

$$-\frac{1}{2}\cos(2x)=-\frac{1}{2}+\sin^2 x=\sin^2 x+C$$
You're right.
Thank you PeroK, Mark44 and Ray Vickson