Integral Calculus Fundamentals

3.3. Terminology. In theintegral b f(x) dx a the numbers a and b are called the bounds of the integral, the function f (x) which is being integrated iscalled the integrand, and the variable x is integration variable. The integration variable is a dummy variable. If you systematically replace it with another variable, the resulting integral will still be the same. For instance, 1 x2 dx= 1 1x3 =1 , 3 3 x=0 0 and if you replace x by you still get 1 1 =1 . 1 3 2 d = 3 3 =0 0 Another way to appreciate that the integration variable is a dummy variable is to look at the Fundamental Theorem again: b f(x) dx= F(b) F(a). a The right hand side tells you that the value of the integral depends on aand b,and has absolutely nothing to do with the variable x. 4. Exercises 336. f(x) = x4 x2 330. What is a Riemann sum of a function y = f(x)? 2 x2 2 x3 3 x4 4 331. Let f be the function f (x) = 1 x. Draw the graph of f (x) with 0 x 2. Compute the Riemann-sum for the partition 337. f(x) = 1 + x+ + + 338. f (x) =1 x 0 < 1 < 1 < 3 < 2 3 339. f(x) = ex 2 of the interval [a, b] = [0, 2] if you choose each ckto be the left endpoint of the interval it belongs to. Draw the corresponding rectangles (add them to your drawing of the graph of f). Then compute the Riemann-sum you get if you choose the ck to be the right endpoint of the interval it belongs to. Make a new drawing of the graph of f and include the rectangles corresponding to the right endpoint Riemann-sum. 340. f (x) =2 x 341. f(x) = e2x 1 342. f(x) = 2 + x ex e x 2 1 1 + x2 ex + e x 2 1 343. f(x) = 344. f (x) = 332. Group Problem. Look at figure1(top). Which choice of intermediate points c1, . . . , c6 leads to the smallest Riemann sum? Which choice would give you the largestRiemann-sum? (Note: in this problem you re not allowed to change the division points xi, only the points ci in between them.) 345. f(x) = 346. f(x) = 1 x2 347. f(x) = sin x Find an antiderivative F (x) for each of the following functions f (x). Finding antiderivatives involves a fair amount of guess work, but with experience it gets easier to guess antiderivatives. 2 348. f(x) = 1 x 349. f(x) = cosx 333. f(x) = 2x+ 1 334. f(x) = 1 3x 350. 351. f(x) = cos2x f(x) = sin(x /3) 335. f(x) = x2 x+ 11 352. f(x) = sin x+ sin 2x 2

353. f(x) = 2x(1 + x2)5 In each of the following exercises you should compute the area of the indicated region, and also of the smallest enclosing rectangle with horizontal and vertical sides. Before computing anything draw the region. 362. The region between the graph of y = 1/x and the x-axis, and between x = a and x = b (here 0 < a <b are constants, e.g. choose a = 1 andb = something against either letter a or b.) 2 if you have 363. The region above the x-axis and below the graph of 1 x 2 354.The region between the vertical lines x = 0 and x = 1, and between the x-axis and the graph of y = x3. 1. f(x) = + 1 + x 355.The region between the vertical lines x = 0 and x = 1, and between the x-axis and the graph of y = xn(here n>0, draw for n= 356.The region above the graph of y = x, below the line y = 2, and between the vertical lines x = 0, x = 4. 357. The region above the x-axis and below the graph of f(x) = x2 x3. 358. The region above the x-axis and below the graph of f(x) = 4x2 x4. 359. The region above the x-axis and below the graph of 4 f(x) = 1 x . 360. The region above the x-axis, below the graph of f(x) = sin x, and between x = 0 and x= . 361.The region above the x-axis, below the graph of f (x) = 1/(1 + x2) (a curve known as Maria Agnesi s witch), andbetween x = 0 andx = 1. 364. Compute 1 1 x2dx 1,1,2,3,4). 2 0 without finding an antiderivative for 1 x2 (you can find such an antiderivative, but it s not easy. This integral is the area of some region: which region is it, and what is that area?) 365. Group Problem. Compute these integrals without finding antideriva- tives. 1/ 2 I = 1 x2dx 0 1 |1 x|dx J = 1 1 |2 x|dx K = 1 5. The indeftnite integral The fundamental theorem tells us that in order to compute the integral of some function f over an interval [a, b] you should first find an antiderivative F of f . In practice, much of the effort required to find an integral goes into finding the antiderivative. In order to simplify the computation ofthe integral b f(x)dx= F(b) F(a) (49) a the following notation is commonly used for the antiderivative: F(x) = f(x)dx. (50) For instance, 1 3 2 3 1 5 sin 5x dx = cos5x, x dx= etc... x , The integral which appears here does not have the integration bounds a and b. It is called an indefinite integral, as opposed to the integral in (49) which is called a definite integral. You use the indefinite integral if you expect the computation of the antiderivative to be a lengthy affair, and you do not want to write the integration bounds aand ball the time. It is important to distinguish between the two kinds of integrals. Here is a list of differences: 3

Indefinite integral Definite integral b af (x)dxis a number. f (x)dx is a function ofx. b af (x)dx was defined in terms of Rie- mann sums and can be interpreted as area under the graph of y = f (x) , at least when f (x) >0. By definition of x whose derivative is f (x). f (x)dx is any function x is not a dummy variable, for example, 2 2xdx = x+ C and are functions of different variables,so they are not equal. x is a dummy variable, for example, 1 2xdx = 1, and 0 1 0 0 2xdx= 2tdt. 1 2 2tdt = t + C 2tdt = 1, so 0 1 5.1. You can always check the answer. Suppose you want to find an antiderivative of a given function f(x) and after a long and messy computation which you don t really trust you get an answer , F (x). You can then throw away the dubious computation and differentiate the F (x) you had found. If Fj(x) turns out to be equal to f (x), then your F (x) is indeed an antiderivative and your computation isn t important anymore. For example, suppose that we want to find ln xdx. My cousin Louie says it might be F(x) = xln x x. Let s see if he s right: d dx x 1 Who knows how Louie thought of this , but it doesn t matter: he s right! We now know that xln x x+ C. . 1 xln x x = x + 1 ln x 1 = ln x. ln xdx= 2. About +C . Let f(x) be a function defined on some interval a x b. If F (x) is an antiderivative of f (x) on this interval, then for any constant C the function F (x) = F (x) + C will also be an antiderivative of f(x). So one given function f(x) has many different antiderivatives, obtained by adding different constants to one given antiderivative. 3. Theorem. If F1(x) and F2(x) are antiderivatives of the same function f(x) on some interval a x b, then there is a constant C such that F1(x) = F2(x) + C. Proof. Consider the difference tt(x) = F1(x) F2(x). Then ttj(x) = F1j(x) F2j(x) = f (x) f (x) = 0, so that tt(x) must be constant. Hence F1(x) F2(x) = C forsome constant. Q It follows that there is some ambiguity in the notation both equal f (x) dx without equaling each other. When this happens, they (F1 and F2) differ by a constant. This can sometimes lead to confusing situations, e.g. you can check that 2sinxcosxdx= sin2x 2sinxcosxdx= cos2x f(x) dx. Two functions F (x) and F (x) can 1 2 are both correct. (Just differentiate the two functions sin2x and cos2x!) These two answers look different until you realize that because of the trig identity sin2x + cos2x = 1 they really only differ by a constant: sin2x= cos2x+ 1. To avoid this kind of confusion we will from now on never forget to include the arbi- trary constant +C in our answer when we compute an antiderivative. 4