Evaluate the following integral, the integral from zero to two of four to the negative Theta, D Theta. So the first thing we have to do is realize that this is a property. To start finding the integral, we're going to look at the derivative. So if we have DDX of A to the X, that's going to equal DDX of E to the lane, A to the X using the power rule bring down the X. So now we know that the derivative of E to something is E to that something times the derivative of the exponent. So we get E XL NA times LNA. If we take the X back up into the exponent using the power rule again, E to the LNA to the X times LNA and then the E to the Ln cancels. So we get a XL NA. Now using the chain rule, we know that we could take the derivative of A to EU where U is a different function. So DDX of A to EU is just going to be A to the ULNADUDX. Thus the integral of A to the UDU is going to be A to EU divided by LNA plus C So when we look at this problem now, we're just really using that formula zero to two, four to the negative Theta, D Theta. Let U equal negative Theta DU equals negative D Theta. Changing our bounds, U of 0 is 0 and U of two is -2, so we're going to go zero to -2 of -4 udu. When we evaluate that by the formula we just talked about, it's going to be -4 / U or -4 U divided by lane 4 from zero to -2, sticking in the -2 for our U and then sticking in the 0. Simplifying, we're going to get 15 / 16 Ln 4. Now there are a couple different formats that could look. You could change the LN4 if you wanted to Ln 2 ^2. So you could have 1530 seconds LN2, but fifteen 16th LN4 works.