# Simple Integral Question

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$\int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt}$

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 $\frac{d^2x}{dt^2}$

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

Edited by Xittenn
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$\int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}\frac{dx}{dt} - \frac{1}{2}\frac{dx}{dt}$

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 $\frac{d^2x}{dt^2}$

Suppose $x(t)=t^2$

$\frac {dx}{dt} = 2t$

$\frac{d^2x}{dt^2} = 2$

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

You have mis-stated something.

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You have mis-stated something.

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

$\int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2$

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

$\int_{a}^{b} \frac{dx}{dt}\frac{d^2x}{dt^2}dt = \int_{a}^{b} \frac{d^2x}{dt^2} dx = \frac{1}{2}x'(b)^2 - \frac{1}{2}x'(a)^2$

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

Or let $u=\frac {dx}{dt}$

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

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