# How do I find the resistive energy deposition of a flat braided aluminum?

I have a flat aluminum 6061, and the measured resistance is 0.5mOhms. I have read that the resistive energy deposition is proportional to the action integral of the lightning current. In the end, I would like to know the amount of power deposited into the aluminum. I was able to find the formula, and it's shown below. How do I find the action integral.

• Substitute the equation for the lightning current waveform into $I$ and perform the integral. – Transistor Jan 1 at 9:58
• Thank you; that helped. – Sam Shurp Jan 9 at 0:19