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(MAR. 2026) A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20-centimeter intervals are 5.8, 20.3, 26.7, 29.0, 27.6, 27.3, 23.8, 20.5, 15.1, 8.7, and 2.8. Use the Midpoint Rule to estimate the area of the wing’s cross-section. =D 200 em —————— A. 4,232 cm? B. 4,333 cm? C. 4,322 cm2 D. 4,222 cm2 woe a0 | J00-Go =o) 149 -100 man] Dx - iaterval of widroiats Midgoint foraula . Area = AX ° | te) + Flin. ) t f(m,) +. + €(my) | Area = 40° [ 20.3 + 22-0 + 27.3+ DS +e | Area & A232 cm* (MAR. 2026) A manufacturer of corrugated metal roofing wants to produce panels that are 60 cm wide and 4 cm high by processing flat sheets of metal as shown in the figure. The profile of the roofing takes the shape of a sine wave. Verify that the sine curve has equation y = sin(1x/7) and find the width w of a flat fetal sheet that is needed to make a 60-cm panel. (Numerically evaluate the integral correct to four significant digits.) w —4en 60cm A. 60.93 cm B. 62.55cm C. 65.22 cm D. 67.89cm (MAR. 2026) The graph of the acceleration a(t) of a car, measured in m/s?, is shown. Use Simpson’s Rule to estimate the increase in the velocity of the car during the 6-second time interval. aah 6 ox = 4 4 2 0 p>! 2445 6 7 (seconds) A. 19.9 m/s B. 16.6 m/s C. 18.8 m/s D. 17.7 m/s al = ~ace> dt JS av = Jat) at Ve Kren of aceelcetime — time Lunctn Simpson's One-Thicd Bale Dens EY] a6 +406 4 20d +406)... 069] ah 6 5 4 O 4 9 3 4 § 6 ft(seconds) Aren = 4 [ ° + 4(0.4) + 2(2) + 4(s) + aod) +444) +0] 3 hren = (2.143 m/s GD Mode- 2-1 (trey ov) x Frea, (0) I 0-4 | 4 2 2 S 4 os | 2 4.7 4 a) | 2x - CA - bepps - $ Sum — Sx cna (1 = 19.193 we 4. (MAR. 2026) The figure shows a circular arc of length s and a chord of length d, both subtended by a central angle 9. Find O > ane) Rw Ow 5. (NOV. 2025) Water is leaking from a tank at the rate of f (t) = 10 In(t +7 ) gallons per hour for 0 (Iy2000) (10-9) ( n¢ * ) 40 C= J (s0- 2K) Ax 20 [C= $400 | 10. (APRIL 2024) Two construction contractors have been hired to clean the Gateway Arch in St. Louis. The Arch is very close to a parabola in shape, 630 feet high and 630 feet across at the bottom. Using the point on the ground directly below the apex of the Arch as the origin, the equation of the Arch is approximately y = 630 — 2 x ys One contractor’s first idea for approaching the project is to build scaffolding in the entire space under 630 the Arch, so that the cleaning crew can easily climb up and down to any point on the Arch. The other contractor A -Z bh thinks there is too much space under the Arch to make (6) the scaffolding practical. To settle the matter, the - & contractors want to find out how much area there is A " 3 (630)(630.) under the Arch. md = = A. 246,600 B. 264,600 | A= Qbacoo C. 244,600 D. 266,600 BIS xo 264600 A= | (a0 -4—) dk \s32-5 -3ls > | = - Is? -S l y 650 Is2-S IS7-SY - 415 — xm (01630 ) x7= 2 ses(y- 630 ) y=o e* V= (0,930) Oo= 11. The work, in joules (J), required to stretch a certain spring a distance of L meters beyond its natural length is given by W = fr 500xdx. How much work is required to stretch the spring 10 centimeters beyond its natural length? A. 5.3J B. 3.5J C. 52J D. 25J 12. (DEC. 2024) The average value of a function f(x) over the interval a s x ee ae eth Cc 14. The volume of a sphere of radius R can be found by slicing the sphere vertically and then integrating the areas of the resulting circular cross sections, this process results in the integral (EGR? — 1x?) dx. this integral to obtain the expression for the volume of a sphere of radius R. LS A. Sart C. tar? 3 EQ ax L=o.lm W = J seox Ax oO as J Kz months fierhlt > ore month (45 + ex- 4x7) dx a. * 420000x Ax 0 I ~-<[ 12 ; Cost: a&l3x* 4 sax[S}+ (Sv +) t &-32x Cost = AB 4 FLAK 4 eR” + 22% east Aer - BLZX~ - dex”? 4+ 6-32 Ax D x = fe no r+@ i) yo Sy = gy = | 280140106m + 8) Cost = By 4 SLZK' 4 Ix” + 622% Ost = 2G + 2G + SO +#22OH Cost = $ 26-4244 G46 17. If the lever in the figure weighs 168 N/m, find its length so as to make the lifting force F a minimum. T.00 | Pir F L-4 5460 N A. 8.602 m B. 8.260m C. 8. 206 m D. 8.062m eo -PLt joe (5) + Styv (1) =0 as -FL + S4l*+S4ded = 0 EAL* +5400 — = . L Ay _ _ _SALO OL L* 18. The efficiency E of a screw is given by: x — px? x+u where u is the coefficient of friction and x the tangent of the pitch angle of the screw. Find x for maximum —— > efficiency if uy = 0.45. A. 0.75 B. 0.85 C. 0.65 D. 0.95 a udy u Van — X= Mx® a ( v ) = v2 F A (x +404 - 24x) - (xe arx?)(£) o- (x4+0-48)(4 7 a(oasya) — (x - o4sx*) | x= 0.WAGS Sul | 21. If the equal sides of an isosceles triangle are Given) U what length of the third side will provide TMaxirr area? A. V3 units B. Cc. v2 units: D. An mim nanayins he x + (FY h V5 units V4 units \l x o} —, nls J N pon Stent nice Nhs . 4 i. Av? wvLto + | \ 4 Me aver A 4 ad “ae ay © ww ) ve ulin we over urine 22. (NOV 2024) Which of the following is/are true? I. Se FG) dx = fy fe) dx + £ f(a) dx Il. Se FG) dx = Sf FR) dx + SP fx) ax Hl. [5 fC) de = ff f(x) de — JP fG) ax WV. fi FG) dx = ff fe) de — J? f(a) ax A. land Il . | only c. il e . band IV £60 de + Fee ax = s Wore Lt) wo Dany ev. |, 23. (NOV 2024) In order to distribute stress, mine tunnels are sometimes rounded. Suppose that the vertical cross sections of a tunnel can be modeled by the parabola y = 6 - 0.06x?. If x and y are measured in feet, bow much rock.would have to be moved to make such a tunnel that is 100 feet long? A. 9,000 ft? B. 7,000 ft? C. 6,000 ft? D. 8,000 ft? 24. Apatch in the shape of the region shown below is to be sewn onto a flag. If each unit in the coordinate system > represents one foot, how much material is required for l % the patch? i ~_ 3x ax A. 10.125 ft? B. 8.125 ft? C. 14.125 ft? D. 12.125 ft? 25. (APRIL 2024) Find the volume generated when the first quadrant area bounded by y = x’, the y axis, and the lines y = 1_and y = 4 is rotated about the y axis in cubic units. y Axis of rotation A. 1717/2 B. 1917/2 Cc. 1511/2 D. 1311/2 27. (NOV. 2023) The first-quadrant area bounded by the curve y = x?, the y axis, and the line y = 4 is rotated about the line x = 3. Use the ring method to find the volume generated. y \ 4 Axis of rotation 28. The deflection y of a simply supported beam with a concentrated load P (see figure) at a distance x from the left end of the beam is given by: = Pbx (L? _ x2 _ b?) ** 6LEI where E is the modulus of elasticity, and | is the moment of inertia of the beam’s cross section. Find the value of x at which the deflecton-s-amaximiur fora eam with a concentrated load 2.00 m —_—. 8.00-m-lo “from the right end. « Lem, Load P 2 Lp? i 2 _ y= | x y A. 4.47 m B. 4.74m Cc. 4.77m D. 4.44m Be - |7- 3x7 b? X= 4.492138485 mM —s<=N 
