5.4 Integration Formulas and the Net Change Theorem [STARTS AT 0:09] INSTRUCTOR: So evaluate the integral from 1 to 4 of root t times 1 minus t Dt. All right, well let's distribute that first. Go from 1 to 4 of root t plus t to the 3 over 2. Go ahead and write this as t to the 1/2. Antiderivative there, again with C equal to 0 is going to be t to the 3/2. And that'll be 2/3 t to the 3/2 dividing by the 3/2. Plus t to the 5 over 2. And that'll be 2/5. Well evaluate that at the end points of 1 and 4. Evaluating at 4 we get 2/3 4 to the 3 over 2, 2/5, 4 to the 5 over 2. Minus evaluating at 1, 2/3. 1 to the 3/2 plus 2/5 1, to the 5 over 2. Which will be 256 over 15, if we take the time to evaluate that. [ENDS AT 1:36]