Lecture for Week 14 Secs 6 3 4 Definite Integrals and the Fundamental Theorem 1 Integral calculus the second half of the subject after differential calculus has two aspects 1 Undoing differentiation This is the problem of finding antiderivatives which we ve already discussed 2 The study of adding things up or accumulating something This is very closely related to the area topic of last week 2 The fundamental theorem of calculus shows that these two things are essentially the same To reveal the basic idea consider a speeddistance problem We know that if an object moves at a constant speed for a certain period of time then the total distance traveled is distance speed time Suppose instead that the speed is a varying function of time If we consider a very short time interval ti 1 ti then the speed is approximately 3 constant at least if f is continuous and hence we can approximate the distance by distance speed in that interval ti ti 1 As the speed we may choose the maximum speed in the short interval or the minimum or anything in between say f wi for some wi ti 1 ti The choice won t make any difference in the end There should be a picture here but I don t have time to draw it now 4 So the total distance traveled between t a and t b is approximately n X f wi ti where ti ti 1 i 1 As we let n the approximation should become exact In effect we have concluded that 5 A The distance traveled between times a and b is the area under the graph of the speed function between the vertical lines t a and t b if area is defined in appropriate units and if the speed is always nonnegative On the other hand B The distance function is an antiderivative of the speed function 6 Conclusion Areas and antiderivatives are very closely related In some sense they are the same thing Before continuing we need a more precise definition of the definite integral as a limit of a sum Z b n X f x i xi f x dx lim a kP k 0 i 1 That occupies the first half of Sec 6 3 and it 7 involves all the issues about varying the sizes xi of the strips etc that I discussed last week about area One important difference between areas and generic definite integrals The integrand function is allowed to be negative in some or all places Areas must always be positive or zero but integrals can be negative In general Z b f x dx A A a 8 where A is the area below the graph and above the x axis and A is the area above the graph and below the x axis In Fig 3 on p 379 A is yellow and A is blue Note also that the dummy variable in an integral can be any letter that doesn t cause confusion Z Z 6 6 x3 dx 2 t3 dt 2 Also this thing is not a function of the variable of integration it is just a number 9 Here is an example where a bad choice of letter would cause confusion 2 x x Z x2 3 t dt 2 must not be written as x2 x Z x2 x3 dx 2 The integral in is a function of x though not of t 10 Algebraic properties of integrals 1 Z b f x g x dx Z b Z f x dx a a 2 Z a b cf x dx c Z b g x dx a b f x dx a 11 for a constant c 3 Z b Z b f x dx a 4 a Z dx means c f x dx a Z Z b f x dx c b 1 dx a and equals b a Remarks Formula 4 is just the area of a rectangle The others are also rather obvious for areas Recall that making 3 obvious was one of the reasons for allowing strips of varying widths xi 12 5 Z a f x dx means b Z b f x dx a if b a It follows that in 3 c does not need to lie between a and b Perhaps more important from the point of view of avoiding mistakes is an identity that is not in the list 13 Z f x g x dx Z f x dx Z g x dx FALSE The integral of a product is not the product of the integrals just as and because the derivative of a product is not the product of the derivatives Roughly speaking every derivative formula turns around to give an integral formula The integral formula corresponding to the product rule for derivatives is integration by parts Sec 8 1 14 Order properties of the integral 1 If f x g x for all x in a b then Z b f x dx a Z b g x dx a Here we assume a b of course In particular if f is nonnegative then Z b f x dx 0 also a 15 2 If f is continuous and f x 0 then Z b f x dx with a b is strictly greater a than 0 unless f x 0 for all x in a b 3 Z b f x dx a Z b f x dx a 16 4 Mean value theorem for integrals If f is continuous then there is a z in a b Rb such that a f x dx f z b a 5 If m f x M in a b then m b a Z b f x dx M b a a 17 Now back to the fundamental theorem Roughly speaking it says that differentiation undoes integration and vice versa They are inverse operations almost like squaring and taking the square root except that they operate on functions instead of numbers The two functions involved are related as are the two readings on a car s speedometerodometer panel The odometer reading is the integral of the speedometer reading The speedo18 meter reading is the derivative of the odometer reading That is the essence of calculus The theorem has two parts one for each order of the operations And I state each part in two versions depending on which function the integral or the derivative takes center stage In stating the theorem we assume for simplicity that f the derivative is continuous 19 Fundamental Theorem Part 1 d dx Z x f t dt f x a That is G x Z x f x dt a is an antiderivative of f 20 Exercise Evaluate d dx Z x 2u du 10 y 1 d dt dy 3 t Z v2 d 2u du dv 10 Z 21 For the first two just apply the theorem Z x d 2u du 2x dx 10 Z y d 1 1 dt dy 3 t y We know these facts without necessarily knowing what the integrals themselves are You may know …
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