This antiderivative is pretty straightforward. Since I have the derivative of 6 5ths t right over here, I can find the antiderivative with respect to 6 5ths t, which is essentially I'm doing u substitution. If you had to do u substitution, you would make that right over there u, and then that, and that would be your du. But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, let me color code this a little bit. When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract.