Evaluate the following integral zero to two of two to the negative Theta D Theta. One way to do this, and there are lots of ways to do it, but one way to do it is to replace 2 to the negative Theta is ELN 2 to the negative Theta. Then bringing out the negative Theta in front as a coefficient would get zero to two of E to the negative Theta LN2D Theta. Now if we did AU substitution, we'd have U equaling negative Theta LN2, IE the exponent. So DU would equal negative LN2D Theta. Remember, LN2 is a constant. If I divide each side by that negative LN2, I'd get -1 divided by LN2DU equaling D Theta. Now going back to the original, we're going to take out the D Theta and we're going to replace it with -1 over LN2DU. And then the E is now going to be to EU because our U was that negative Theta L and two. We need to change our bounds. So if we want to figure out what U of 0 is, we have -0 * l and two, which is 0, and then we're going to put in U of two, so -2 L and two. So now we get a new integral of -1 / l and two from zero to -2 L and two E to the UDU. And we know that E to the UDU, the integral is just E to EU. So now we stick in our upper bound minus our lower bound E to the -2 L and two. I can take that -2 back up into the exponent and I get E to the lane of two to the -2, which is just a fourth. Any E to the zero, anything to the zero power is one other than 0 to the zero. So if I have 1/4 -, 1, that gives me a negative 3/4. If I multiply -3 fourths times the -1 over lane two, we can see that that's 3 / 4 lane 2.