Calculus/Fundamental Theorem of Calculus

Definite integral Indefinite integral The fundamental theorem of calculus is a critical portion of calculus because it links the concept of a derivative to that of an integral. As a result, we can use our knowledge of derivatives to find the area under the curve, which is often quicker and simpler than using the definition of the integral. As an illustrative example see § Hyperbolic logarithm and angles for the connection of natural logarithm and 1/x. Mean Value Theorem for Integration We will need the following theorem in the discussion of the Fundamental Theorem of Calculus. Mean Value Theorem for Integration Suppose is continuous on . Then for some . Proof of the Mean Value Theorem for Integration satisfies the requirements of the Extreme Value Theorem, so it has a minimum and a maximum in . Since and since for all we have Since is continuous, by the Intermediate Value Theorem there is some with such that Fundamental Theorem of Calculus Fundamental theorem of calculus Statement of the Fundamental Theorem Suppose that is continuous on . We can define a function by Fundamental Theorem of Calculus Part I Suppose is continuous on and is defined by Then is differentiable on and for all , When we have such functions and where for every in some interval we say that is the antiderivative of on . Fundamental Theorem of Calculus Part II Suppose that is continuous on and that is any antiderivative of . Then thumb|350px|right|Figure 1 Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus. Proofs Proof of Fundamental Theorem of Calculus Part I Suppose . Pick so that . Then and Subtracting the two equations gives Now so rearranging this we have According to the Mean Value Theorem for Integration, there exists a such that Notice that depends on . Anyway what we have shown is that and dividing both sides by gives Take the limit as we get the definition of the derivative of at so we have To find the other limit, we will use the squeeze theorem. , so . Hence, As is continuous we have which completes the proof. Proof of Fundamental Theorem of Calculus Part II Define . Then by the Fundamental Theorem of Calculus part I we know that is differentiable on and for all So is an antiderivative of . Since we were assuming that was also an antiderivative for all , Let . The Mean Value Theorem applied to on with says that for some in . But since for all in , must equal for all in , i.e. g(x) is constant on . This implies there is a constant such that for all , and as is continuous we see this holds when and as well. And putting gives Notation for Evaluating Definite Integrals The second part of the Fundamental Theorem of Calculus gives us a way to calculate definite integrals. Just find an antiderivative of the integrand, and subtract the value of the antiderivative at the lower bound from the value of the antiderivative at the upper bound. That is where . As a convenience, we use the notation to represent Integration of Polynomials Using the power rule for differentiation we can find a formula for the integral of a power using the Fundamental Theorem of Calculus. Let . We want to find an antiderivative for . Since the differentiation rule for powers lowers the power by 1 we have that As long as we can divide by to get So the function is an antiderivative of . If then is continuous on and, by applying the Fundamental Theorem of Calculus, we can calculate the integral of to get the following rule. Power Rule of Integration I as long as and . Notice that we allow all values of , even negative or fractional. If then this works even if includes . Power Rule of Integration II as long as . Examples To find we raise the power by 1 and have to divide by 4. So The power rule also works for negative powers. For instance We can also use the power rule for fractional powers. For instance Using linearity the power rule can also be thought of as applying to constants. For example, Using the linearity rule we can now integrate any polynomial. For example Exercises 1. Evaluate . Compare your answer to the answer you got for exercise 1 in section Definite integral. 2. Evaluate . Compare your answer to the answer you got for exercise 2 in section Definite integral. 3. Evaluate . Compare your answer to the answer you got for exercise 4 in section Definite integral.4. Compute 5. Evaluate . 6. Given , then find . 7. Let . Then find the .8. Given . Then find .9. If . Then find . 10. For the function over the given closed interval, , find the value(s) of guaranteed by the mean value theorem for the definite integral. Solutions Definite integral Indefinite integral