1. 4.4 The Fundamental Theorem of Calculus and The Second Fundamental Theorem of Calculus

2. The fundamental theorem of calculus explains how differentiation and (definite) integration are inverse operations. Similar to how multiplication and division are inverse operations.

3. The Fundamental Theorem of Calculus If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then Which means that to evaluate a definite integral, we can find the integral of f and evaluate that expression @ a, and then evaluate it @b and then subtract them. It is f evaluated at b minus f evaluated at a.

4. This allows us to evaluate an integral without using the limit of a sum. We will use the following notation.

5. You do not have to include your constant of integration C in these problems because of the following idea.

6. Evaluate the following

12. -When the graph goes below the x-axis we run into the problem that we get back a negative area. When this happens we need to pay attention to our answers and what the question or application is asking for. -If it wants Net Area you just complete the integration, in this case some parts will cancel out other parts. -If it wants total area then you need to pay attention to what parts are negative and what parts are positive. -If it wants you to evaluate the definite integral or to find the definite integral simply use the fundamental theorem of calc and ignore anything about the function being below the x-axis. -For the most part, in this text, it mainly asks you to evaluate the integral. When the curve falls below the x-axis

13. 1

14. 2 If this is the method we want to use, then we simply break the integral up into 2 integrals with different upper and lower limits, to see where the graph crosses the x-axis simply set y=0 to find the x-intercepts.

15. If we do not pay attention to these 2 cases we could have a situation like the following. Find the area between the x-axis and the curve y=x 3 integrated between -2 and 2. There is obviously area present, but the areas are both congruent regions. One is positive, one is negative. So if we integrate between -2 and 2, we would get 0. So we need to integrate between -2 and 0, (result is a negative) take its absolute value, then integrate between 0 and 2. Then sum the values.

16. These explanations have been given because of their use in certain applications. Know that when we take the definite integral using the Fundamental Theorem of Calculus “the definite integral of a continuous function over an interval is the sum of the areas above the x-axis minus the sum of the areas below the x-axis”

17. Homework pg. 291 5-13

18. The Second Fundamental Theorem of Calculus We have seen that the definite integral of f on the interval [a,b] was defined by the constant b as the upper limit of integration and x as the variable of integration. A different situation may arise in which the variable x is used as the upper limit of integration, if this is the case then t is usually used as the variable of integration.

19. Evaluate the function We could substitute the values of x into the integral and integrate and evaluate 4 separate times, but to be more efficient there is a better method. Integrate the function with the lower limit 0 and the upper limit x. ----------------------------------------------------------------------------------------------- Now plug in the values of x into this function.

21. ACCUMULATION FUNCTION We can think of this idea as t goes from 0 to x, as an accumulation function, we can slide x around and find out how the area of the function changes as x changes. This is a critical idea in many real-world applications of integration.

22. Evaluate the following