The place of the elementary calculus in the senior high-school mathematics : and suggestions for a modern presentation of the subject . erentials of the following functions. f(x) = 5X2. Then df(x) = loxdx. f(t) = 4^ - 2>t2 - 1. Then df(t) = \2tHt - 6tdt. Find the differentials of the following: 5*2. 4/3- 3t2~ ~ I, av2 — b, 72s - z, kt, y2gt, hta - 2t - - 3, cxn~ 1 > Ans. loxdx. Ans. I2t2dt - 6tdt. Ans. 2avdv. Ans. 2\zLdz — dz. Ans. kdt. Ans. y2gdt. Ans. ahf ~ ldt - 2dt. Ans. (n-i)cxn~2. The pupils should solve many additional examples of thiskind. INTEGRATION A Reason for Studying Integr
The place of the elementary calculus in the senior high-school mathematics : and suggestions for a modern presentation of the subject . erentials of the following functions. f(x) = 5X2. Then df(x) = loxdx. f(t) = 4^ - 2>t2 - 1. Then df(t) = \2tHt - 6tdt. Find the differentials of the following: 5*2. 4/3- 3t2~ ~ I, av2 — b, 72s - z, kt, y2gt, hta - 2t - - 3, cxn~ 1 > Ans. loxdx. Ans. I2t2dt - 6tdt. Ans. 2avdv. Ans. 2\zLdz — dz. Ans. kdt. Ans. y2gdt. Ans. ahf ~ ldt - 2dt. Ans. (n-i)cxn~2. The pupils should solve many additional examples of thiskind. INTEGRATION A Reason for Studying Integration. One of the reasons for study-ing integration is that it enables one to find the areas of figures whichcannot be found by the ordinary geometric methods. Plot the curve/ (x) = x2 — \x + 5- Erect ordinates at x = a and x = b. Howcan the area enclosed by these ordinates, the curve, and X-axis befound? This will now be shown. 14 The general treatment of the differential of which the method here given is anoriginal adaptation is found in Granvilles Calculus, p. 132. 74 Elementary Calculus in Senior High-School Mathematics. At the point P the abscissa x equals a. Give x a small this by dx. This may be done since under the topic of the differential it was pointed out thati \f(X)-qxis , the increment of the independent variable x equals dx. At the right-hand extremity of dx erect an or-dinate. Then this small strip ofthe area is almost rectangular inshape with a base equal to dx andan altitude approximately equalto the ordinate / (x). Then thearea of this strip is approxiatelyf (x)-dx. Next divide the remain-der of the area into strips eachwith its base equal to dx. Let x assume these different points ofdivision considered as abscissas along the X-axis. Then for each ofthese values of x the corresponding values of the ordinates will be/ (x), and the area of each of these strips will be represented by/ {x)-dx. The sum of all such strips is approximately
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