How do I integrate i and L?

[vorcil]

I wanted to see the proof for the energy stored in an inductor equation was,
But i had trouble understand how part of this integration works i.e my math sucks

-

given $$\frac{dw}{dt} = iv$$

and $$VL = L \frac{di}{dt}$$

-

solving i get,
subtituting the ldt/dt for v

$$\frac{dw}{dt} = i L \frac{di}{dt}$$

the dt's cancel out

$$dw = i L di$$

then to find the work done I integrate the equation,

$$\int dw = \int i L di$$

and the integral of $$\int dw = w$$

but how do I integrate

$$\int i L di$$ ?

I know L, the inductance of the inductor is constant so can pull that outside the integral,
and get
$$L \int i di$$

But what do I do here?

integrating i I get $$\frac{1}{2} i ^2 * i$$ according to the integration rules I've learnt

but everyone knows that the energy inside an inductor equation is

$$wL = \frac{1}{2} L i^2,$$

but my integration shows it is $$\frac{1}{2} L i^3$$

can someone please explain it to me

 

[rock.freak667]

First I think this equation

$$VL = L \frac{di}{dt}$$

should be

$$V=L\frac{di}{dt}$$

Well you got as far as W= L∫i di.


Now remember the rules of integration

$$\int x^n dx = \frac{x^{n+1}}{n+1}$$

so what does your integral become?

 

[vorcil]

when I wrote VL i ment the voltage across the inductor Vl,

$$\int i^1 di = \frac{i^(1+1)}{1+1}$$

I just always assumed that the integral of something $$\int di = i$$
or $$\int dx = x$$

since when I integrate, $$\int dw = L \int i di$$
I get $$w = \frac{1}{2} L i^2$$

-

I can see now that the di disappears, and all I'm integrating is i, which I can do easily

 

[rock.freak667]

I just always assumed that the integral of something $$\int di = i$$
or $$\int dx = x$$

yes those are correct, but you were integrating i w.r.t i. Not just 1.

 

[vorcil]

yes those are correct, but you were integrating i w.r.t i. Not just 1.

so when I integrate x w.r.t x, I get the integral of x only?

$$\int x dx = \int x = \frac{1}{2}x^2$$

but when I integrate dx,

$$\int dx = x$$ ?

so that dx after an equation means with respect to dx?
like in an integral equation, $$\int x dx$$ is means integrating x with respect to x?

 

[rock.freak667]

so when I integrate x w.r.t x, I get the integral of x only?

$$\int x dx = \int x = \frac{1}{2}x^2$$

but when I integrate dx,

$$\int dx = x$$ ?

so that dx after an equation means with respect to dx?
like in an integral equation, $$\int x dx$$ is means integrating x with respect to x?

Yes. But the 'dx' does not just disappear.

$$\int 1 dx = x+C$$


$$\int x dx = \frac{1}{2}x^2 +C$$

(constant C for indefinite integrals)

 

[vorcil]

ok thanks, I'll figure out how differential/integral equations are set up properly

instead of just using my cheap way and plugging everything into

$$\int x^n dx = \frac{x^{n+1}}{n+1}$$

 

[Mark44]

The differential part in an integral tells you what the variable of integration is. Don't leave it out as you did here. Also, don't leave out the constant of integration.

$$\int x = \frac{1}{2}x^2$$

It should be
$$\int x dx = \frac{1}{2}x^2 + C$$

In this integral, a is the variable of integration.
$$\int a da = \frac{1}{2}a^2 + C$$

In this one, x is the variable of integration, and a is just a constant.
$$\int a dx = a \int dx = ax + C$$
In the integral above I used a property of integrals - the integral of a constant times a function is the constant times the integral of the function. Here the function is 1.