By D. J. Bell and F. H. George (Auth.)

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62 - 2 1 f t 65 6 ft. 39 21ft. 59 from 66 The area MNQP is greater than the area of the rectangle MNSP but smaller than the area of the rectangle MNQR. To put this into mathematical terms, suppose P is the point (x, y) and Q the point (x + δχ, y + ôy). Let area ABDC be denoted by S and area MNQP by OS. Then y. δχ < öS < (y + ôy). <5x. Since δχ is positive it follows that y<-

Ii Turn to 56 40 from 32 Your answer was "motion along a straight line with zero acceleration". This was a wrong answer. Do not forget that for uniform velocity the magnitude and the direction of the velocity have to be constant. This is true in the case of motion in a straight line with zero acceleration since zero acceleration implies constant speed. Turn back to 32 41 from 31 Your answer was 512 ft/sec. This was a wrong answer. You have calculated the distance travelled in 4 sec, since 8i3 = 512 when t = 4.

The line PS is drawn to complete the rectangle MNSP and, similarly, RQ completes the rectangle MNQR. Similar constructions are carried out for all the strips between AC and BD. Because we shall be referring to the above diagram again, you should make a copy of it on a sheet of paper. Turn back to 59 67 from 53/63 From the formula dS it follows that the area S under the curve between x = a and x = b is the definite integral $ydx = a S(b)-S(al where S(x) denotes the indefinite integral of y with respect to x.