CHAPTER VII INTEGRATION 105. Introduction. In Chapter VI we considered the problem: Given the function y, to find the derivative Day. In this Chapter we shall consider the inverse problem: Given the derivative Day, to find the function y. DEFINITION. A function whose derivative is a given function is called an integral of the given function and the process of finding it is called integration. Thus, if Daf(x) = F(x), then f(x) is an integral of F(x). EXAMPLE 1. Given Day = 2x, find y. = Since Dxxnxn−1, we have Dxx2 = 2x. Hence r2 is an integral of 2x. But the derivative of each of the functions x2 + 5, x2 7, x2+, is 2x. Hence these functions are also integrals of 2x. In fact, any function of the form y x2 + C, where C is any constant, has the derivative 2x, and is therefore an integral of 2x. = Thus we see that the process of differentiating the function x2 gives rise to the single derivative 2x, but that the inverse process of integrating 2x gives rise to an indefinite number of integrals x2 + C which differ by a constant. Theorem. If two functions f(x) and F(x) have the same derivative, their difference is a constant. But a function whose derivative is equal to zero for all values of the variable is a constant. For the tangent line at every point on the graph is horizontal, so that the graph of the function is a straight line parallel to the x-axis. y = f(x) − F(x) = C, f(x) = F(x) + C. That is, every integral can be obtained from a given integral by adding the proper constant. The constant C which is added to a known integral of a function is called the constant of integration and it should not be omitted in performing the operation of integration. If y is a function of x whose derivative with respect to x is given, say F(x), then Dxy = F(x). The notation employed to indicate that y is an integral of F(x) is SF(x)dx, the equation being read "y equals the integral of F(x)dx." In the illustration above, since Day 2x, then = S2x In a particular problem sufficient data are usually given to determine the constant of integration and to enable us to find a particular integral which satisfies the given conditions. The graphical significance of the constant of integration and the method of determining its numerical value so as to satisfy given conditions is shown in EXAMPLE 2. Find y if Day = x2 - 4 and Find the interpret the result graphically. equation of that one of the resulting curves which passes through the point (1, 2). By differentiating 3 we obtain 3x2, and hence we must differentiate in order to get x2. Differentiating 4x gives 4, and hence if Dxy = x2 - 4, where C is the constant of integration. y = 3x3 − 4x + C, This represents a set of curves which may be obtained from one of them by moving it up or down (Theorem, page 19). Several curves of this set are shown in the figure. These curves have parallel tangent lines at points with the same abscissas, since the derivative of y with respect to x is the same for each function, namely, Dxy = x2 - 4. The intercepts of these curves on the y-axis are the respective values of C. If (1, 2) lies on one of these curves then 2 = · 13 – 4.1 + C, whence C = 53 and the required equation is y x3- 4x + 53. = EXAMPLE 3. If Dxy = ax”, show that y = a -xn+1 + C. n+1 The result is correct because if we differentiate it we get Hence, to integrate a term of the form ax", increase the exponent by one and divide the coefficient by the new exponent. In like manner, if Day = au" D2u, then y a = un+1 + C. n+ 1 If the integral notation is used, these results may be written = x3 + 4x2 + C. (b) Integrate S (3x2 + 8x)dx. We have (3x2 + 8x)dx = ƒ3x2 dx + S8x dx (c) Find y, if Dxy = §(x2 + 3x)1 (2x + 3). If we think of x2 + 3x as a function u, then 2x + 3 is Dau, and therefore Differentiation is a direct process while integration is an inverse process. The former can be carried out according to a general method of procedure but there is no general method of procedure in the case of the latter. Integration in the last analysis is a matter of trial in which attempts are made to reduce the given function to a form the integral of which is known. A thorough knowledge of differentiation is essential for the process of integration. EXERCISES In Exercises 1–12, find y if Day is the given function, and illustrate the result graphically. Determine the equation of that member of the set of curves which passes through the indicated point. 13. For what value of n does the result in Example 3 above fail? = (x2 + 3x − 2)2 (2x + 3), find y. 14. Given Day 16. Find the equation of the curve whose slope at any point is 2x + 3 and which passes through the point (2, 3). Find the equation of the tangent line at this point. Plot the curve. 17. The slope of a curve at any point is 6x2 + 4x and it passes through the origin. Find the equation of the curve and the equation of the normal line at the point of inflection. Plot the graphs of the two equations. 18. Find the cubic function whose graph has a maximum point at (1, 4) and a minimum point at (3, – 5). Find the equation of the tangent line at the point of inflection. Suggestion. Day = a(x − 1)(x − 3). 19. The slope of a curve at any point is 4x - 3 and it passes through Find the equation of the curve and the coördinates of the minimum point. Plot the graph. the point (1, 2). Since A changes as x changes and is determined when x is fixed, A is a function of x. As N moves to R, x increases by an amount Ax, A by an amount PNRQ = AA, and y by an amount SQ = Ay. Since PNRQ is less than the rectangle TNRQ and greater than the rectangle PNRS, we have As Ax approaches zero, y + Ay approaches y and AA/Ax ap ΔΑ proaches DA; and as lies between the magnitudes y and Ax Theorem. The rate at which the area A changes with respect to x is equal to the right-hand ordinate of the bounding curve. Symbolically, The area A can now be found as a function of x by integration, the constant of integration being determined by the fact that A = 0 when x = a. If a fixed right-hand boundary is chosen the area is determined. The method of finding the area is shown in EXAMPLE 1. Find the area bounded by the curve y = x2, the x-axis and the ordinates at x = 1 and x = = 4. $21 18+ B 15 12 -9 Equation (4) gives the area under the curve starting at the ordinate MP, at x = 1, and continuing to any second ordinate. In this case the second ordinate is fixed at x = 4. Hence substituting x = 4 in (4) we have |