The graph of (4) is the curve LMD which crosses the x-axis at x = 1 where the area MPBN begins. The number of square units in MBPN is the same as the number of linear units in the ordinate ND, which is 21 units in length. Equation (1) may be interpreted thus: Suppose that we had y = c, whose graph is a straight line parallel to the x-axis. Then the rate of change of A with respect to x, DA, would be uniform, since DA y = c. It would be measured by the change in A due to a unit change in x (page 48), that is, by a rectangle with base unity and altitude c. B M A C N FIG. 176. 1 = Returning to the figure used in proving the theorem, let NR = 1. Then the area of the rectangle PNRS y x1 = y is the amount by which A would increase when x increases by unity, provided that A increased uniformly for x>ON. = The method of proving the theorem above may be used to show that an area A bounded by a curve, the y-axis, and two abscissas (one fixed, the other not), is such that 1. Find the area under each of the following curves from x = = 4. 2. Find the area bounded by the curve y = x2, the x-axis and the line x = 2, and the area bounded by the curve, the y-axis and the line y Check by finding the area of the rectangle formed by the axes and the lines x = 2 and y = 4. = 3. Find the area above the x-axis and below the parabola y 4. If the positive intercepts of the parabola y coördinate axes are denoted by A and B, and if a tangent parallel to the chord AB cuts the ordinates at A and B in C and D, and touches the parabola in E, show that the area of the parabolic segment ABE is twothirds that of the parallelogram ABCD. = 5. If the tangent to the parabola y 3x2 - 6x+8, which is parallel to the chord joining the minimum point A to the point B on the curve whose abscissa is 2, cuts the ordinates of A and B in C and D, find the relation between the area of the parabolic segment from A to B and the area of the parallelogram ABCD. 6. Find the area bounded by the following pairs of curves: = 7. Find the area bounded by the curve y x3 tangent to the curve at the point of inflection, and the ordinates at the maximum and minimum points. 8. Show that the area under the parabola y = ax2 + bx + c between (Y1 + 4Y2 + Y3), where y1, Y2, Y3 k respectively. Suggestion: Sub stitute the values of y in the above expression and show that the result is the same as the area under the curve. 9. Find the area under the following parabolas and check the results by means of the formula for the area given in Exercise 8. 10. Show that the area formula in Exercise 8 is true for any cubic function y ax3 + bx2 + cx + d. = = 11. Find the area under the curve y x3 + 12x + 4 from x = Check the result by means of the formula in Exercise 8. x 1 to x = 3. DEFINITION. The average ordinate of a curve y = f(x) from = a to x = b is the height of the rectangle with base b – a, whose area is equal to the area under the curve from x = a to x = b. If y represents the average ordinate and A the area under the curve from x = a to x = b, then 12. Find the average ordinate of the following curves for the ranges indicated. = 13. Given t3 — 9t2 + 15t + 30, where is the temperature at any time t, find the average temperature for the range from maximum to minimum temperature. 14. The pressure p and the volume v of a gas are connected by the equation pv1.2 = k. and p = 15 pounds per square inch when v = 0.5 of a cubic foot. Find the average pressure of the gas in expanding from 1 to 3 cubic feet. 15. The tension T of a spring is connected with the amount of stretching s by the equation T = 1⁄2s + 3. Find the average tension as s changes from 0 to 3. 16. Find the percentage of error in the area under the curve y = x2 - 2x + 3 from x = · 0 to x = 2, due to an error of 1 per cent in the range. 107. Motion in a Straight Line. The fundamental notions involved in the motion of a particle in a straight line are: The distance, s, measured from a convenient station on the line. The time, t, which has elapsed from a fixed time. The motion may be described by an equation involving two or more of the magnitudes t, s, v, a. We have already considered motions described by an equation of the form s = f(t), and obtained the velocity and acceleration by differentiation. We shall now consider two other types in which integration is involved. If the velocity is given as a function of the time by an equation of the form v = ƒ(t), the acceleration, a, is found by differentiating while the space, s, is found by integrating this equation. The constant involved in the integration is determined as in the following example. EXAMPLE 1. A body moves from rest in a straight line so that its velocity after t seconds is given by the law v = 4t. Assuming that a ball placed on a plane inclined at arc sin would roll according to this law, determine: (a) How far it will roll in 2 seconds. (b) Its distance from the upper end of the plane after 1.5 seconds, if it is placed 4 feet from that end. (c) Its distance from the lower end after 1 second, if it is placed 12 feet from the lower end. We know that if s represents distance from a certain point at the time t, then Dis = v, and hence, in this example, Dis= 4t. The required function, s, is therefore one whose derivative with respect to t is 4t. One such function is 2t2, and adding the constant of integration (page 302), we have The value of the constant of integration may be determined if we know a pair of values of t and s, and these depend on when and where the ball starts to roll. We shall assume that t O when the ball begins to roll. The distance s is measured from some chosen station O. = (a) The distance the ball will roll in 2 seconds is the same as its distance from the starting point after 2 seconds. Hence for this part of the problem we choose the station O at the starting point, so that s O when t=0. Substituting these values in (1) we find that C = 0. Hence, substituting this value of C in (1), the distance from the starting point after t seconds is Then at the end of 2 seconds, 8 (b) Take O at the upper end from O, we have s = 4 when t S = 212. = 2 x 228 feet. = = (2) of the plane. Since the ball starts 4 feet 0. Substituting these values in (1), we find that C = 4. Setting this value of C in (1), the distance from the upper end of the plane to the ball after t seconds is That is, the ball is 8.5 feet from the upper end of the plane after 1.5 seconds. (c) If O is taken at the lower end of the plane and the ball placed 12 feet above it, then s = - 12 when tC (s being negative since the positive direction for s is the same as that of v, namely, down the plane). Substituting in (1) we find C= 12, so that the distance from the lower end of the plane at any time is If t = 1, 8 = C-4 -18 2 3 t - 10, that is, after one second the ball is 10 feet above the lower end of the plane. Graphically, the given relation v = sented by the straight line through the origin whose slope is 4. The graph of (1) is given for C = 0, 4, - 12, which are the values of C in (2), (3), (4), respectively. The intercepts of these curves on the s-axis are the respective values of C, which give the distances from the station O to the ball when t If the acceleration is given as a function of the time, t, by an equation of the form a = f(t), the velocity is obtained by integrating this equation and the distance by integrating the result. Two constants of integration are introduced in the process which are determined by given conditions as in EXAMPLE 2. A balloon is ascending with a uniform velocity of 28 feet per second and at the height of 720 feet a ball is dropped. When will the ball strike the ground and with what velocity? = Let s be the height of the ball above the ground, and let t O when the ball is dropped. The direction of the acceleration of gravity, being toward the center of the earth, is opposite to the positive direction of s, and hence the acceleration is negative. The initial conditions and required data are collected in the table. As soon as the ball leaves the balloon it is subject to the law When the ball strikes the ground s = 0. Substituting this value of s in (5) the negative value of t having no meaning in this problem. Substituting this value of t in (3) 1. A ball is rolled up a plane inclined at an angle of 10° with an initial velocity of 15 feet per second. Find the distance it rolls up the plane and the time that elapses before it returns to the starting point. (Suggestion: Find the component of the acceleration of gravity acting along the plane.) |