Simple Integral Question

[Tracker]

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt} [/math]

 

What I am having trouble is understanding is going from the second step to the third step. I don't understand how you can integrate with respect to dx and still integrate [math] \frac{d^2x}{dt^2} [/math]

[Xittenn]

[math]

 

\frac{d^2x}{dt^2}=\frac{d}{dt} \frac{dx}{dt}=v \frac{d}{dt} = \frac{dv}{dt} = \frac{dv}{dt} \cdot \frac{dx}{dx}

 

[/math]

Edited February 26, 2011 by Xittenn

[DrRocket]

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt} [/math]

 

What I am having trouble is understanding is going from the second step to the third step. I don't understand how you can integrate with respect to dx and still integrate [math] \frac{d^2x}{dt^2} [/math]

 

Suppose [math] x(t)=t^2[/math]

 

[math] \frac {dx}{dt} = 2t [/math]

 

[math]\frac{d^2x}{dt^2} = 2 [/math]

 

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}\ dt = \int_a^b 4t \ dt[/math] [math]= 2b^2-2a^2[/math] [math] \ne\frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt} [/math] [math]=0[/math]

 

You have mis-stated something.

[Tracker]

 

You have mis-stated something.

 

 

The third part was wrong. Here is the correct version:

 

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2 [/math]

[DrRocket]

The third part was wrong. Here is the correct version:

 

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2 [/math]

 

Look at [math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt[/math] and integrate by parts.

 

Or let [math]u=\frac {dx}{dt}[/math]

 

[math] \int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{x'(a)}^{x'(b)} u \frac{du}{dt} dt[/math] [math] = \int_{x'(a)}^{x'(b)} u du = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2 [/math]