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We observe that the expression 6x 2 + 4x is nearly the derivative of x 3 + x 2 + 1 In fact if we set ( x) = x 3 + x 2 + 1 then the integrand (the quantity that we are asked to integrate) is [ ( x) ]50 2 ( x) It is natural to guess as our antiderivative [ ( x) ]51 Checking our work, we find that [ ( x) ]51 = 51[ ( x) ]50 ( x)

310000 320000 330000 340000 350000 360000 370000 380000 390000 400000 410000 420000 430000 440000 450000 460000 470000 480000 490000 500000 510000 520000 530000 540000 550000 560000 570000 580000 590000 600000 610000

(Continued)

We see that the answer obtained is quite close to the answer we seek; it is off by a numerical factor of 2/51 With this knowledge, we write our final answer as ( x 3 + x 2 + 1) 50 ( 6x 2 + 4x) dx = 2 [x 3 + x 2 + 1]51 + C 51

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CAMSHAFT ANGLE (DEGREES) PRESSURE ANGLE (DEGREES) 253656 267994 281369 293788 305262 315813 325464 334244 342181 349308 355657 361259 366146 370347 373893 376810 379126 380864 382049 382702 382843 382491 381663 380376 378644 376480 373898 370909 367523 363751 359602 14565 14690 14824 14967 15118 15278 15446 15622 15806 15997 16196 16402 16614 16832 17057 17287 17523 17764 18009 18258 18511 18768 19028 19290 19554 19820 20088 20356 20625 20894 21162 43388 46624 49812 52943 56013 59014 61939 64783 67538 70200 72762 75219 77566 79797 81907 83893 85749 87472 89058 90503 91805 92960 93966 94822 95524 96072 96464 96700 96778 96700 96464 1172860 1156628 1137891 1116689 1093069 1067081 1038783 1008235 975503 940658 903776 864937 824224 781727 737536 691748 644462 595780 545807 494653 442427 389243 335217 280464 225103 169255 113041 56582 00000 -56582 -113041

Consider the curve shown in Figure 41 The curve is the graph of y = f (x) We set for ourselves the task of calculating the area A that is (i) under the curve, (ii) above the x-axis, and (iii) between x = a and x = b Refer to Figure 42 to see the geometric region we are considering We take it for granted that the area of a rectangle of length and width w is w Now our strategy is to divide the base interval [a, b] into equal subintervals Fix an integer k > 0 We designate the points P = {x0 , x1 , x2 , , xk } , with x0 = a and xk = b We require that |x j x j 1 | = |b a|/k x for j = 1, k In other words, the points x0 , x1 , , xk are equally spaced We call the

FOLLOWER ANGLE (DEGREES)

+

CAMCO B-12345 (SAMPLE) FOLLOWER RADIUS FOLLOWER VELOC FOLLOWER ACCEL (INCHES) (INCHES/SEC) (INCH/SEC/SEC) -33453 -42677 -58063 -88364 -173939 -1857170 232065 113671 77297 59720 49406 42652 37902 34391 31696 29567 27843 26421 25227 24209 23330 22561 21882 21275 20727 20229 19772 19349 18955 18586 18239

y = f (x)

RADIUS OF CURV (INCHES)

620000 630000 640000 650000 660000 670000 680000 690000 700000 710000 720000 730000 740000 750000 760000 770000 780000 790000 800000 810000 820000 830000 840000 850000 860000 870000 880000 890000 900000 910000 920000

y = f (x)

620000 630000 640000 650000 660000 670000 680000 690000 700000 710000 720000 730000 740000 750000 760000 770000 780000 790000 800000 810000 820000 830000 840000 850000 860000 870000 880000 890000 900000 910000 920000

I Writing ds =-(x2 +y2)-l12d(x2 y 2 ) + 2 as the first step will help to prevent errors with 1/2's and the minus signs

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