2016 FAMAT Convention Mu Integration 1 = 80 0 = 80. dx 1 + x 2 = arctan x] k2

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1 6 FAMAT Convention Mu Integration. A ] B. For quadratic functions, Simpson s Rule is eact. Thus, 3. D.. B. lim ) 3 + ] cot θ) dθ csc θ ) dθ cot θ θ + C n k n + k n lim n k n + k n + arctan ] 5. E. Using the Fundamental Theorem of Calculus, the derivative of the error function d erf) e. It is clear that the derivative only takes on positive values on its domain, so the error function must be monotonically increasing. The second derivative of the error function d erf) e. The second derivative takes on both positive and negative values on its domain so the error function is not eclusively concave up or concave down on its domain. 6. C. sin + tan sec ] lim b sin + sin ) sin ) tan b sec b + lim b sin b cos b sin cos + lim b sec sec tan ) cos b sin b C. While this problem is doable in its present state, knowledge of hyperbolic functions makes e + e it far easier. cosh and e e sinh. By the Disk/Washer Method, the volume is equal to cosh sinh ). The identity cosh sinh makes this very simple: 8. C. ] sin 5 ) cos ) cos ) tan sin cos ) sec ) sin sec + cos ) sin sec tan + cos sin sin ) sec 3 cos3 + cos + C

2 6 FAMAT Convention Mu Integration 9. E. lim n ) n ) ] 7. C. b a y sin θ ) sin θ dθ). Because of the periodicity of sin 3 θ, this integral is equal to zero and we are looking for the area, not just the integral.. B.. A. sin θ ) sin θ dθ) cos θ ) sin θ) dθ cos θ sin θ sin θ ) dθ [ cos θ 3 cos3 θ ] becomes 3. B. β R 8 3 e 3 + e +. Let u e so the integrand has u3 + u + u u +. Then the integral e e + ) e e + ] e e + + e e + 3 ) dr β β r + dθ sin θ + cos dθ θ dθ dθ β arctan 3] 6 ) 6 3. C. Let Q be the total charge carried by the sphere and V be the volume of the sphere. So, for infinitesimal values dq and dv, ρ dq. So dq ρ dv dq ρ dv. For a dv 5kr sphere, V 3 r3 dv r dr. So dq ) r dr ). Let s simplify and 5. E. insert our limits: Q dq R sec 3 6kr dr Q kr 5] R kr5 sec sec u sec, dv sec sec 3 sec tan sec tan sec tan sec sec ) sec tan + sec sec 3 sec 3 sec tan + ln sec + tan sec 3 sec tan + ln sec + tan + C

3 6 FAMAT Convention Mu Integration 3 Both the original function sec 3 ) and its antiderivative have a singularity at integral diverges. so the 6. A. arcsin) arcsin ) arcsin ) arcsin ) arcsin) + arcsin ) + arcsin) ] ) 7. C. y y y y y ± +. But we want the positive value, so y + + and ) ) 3 ] D. Integral test: diverges. ln Thus, B converges. ln u ln, du u ln, dv ln + du u ln u] ln ln ln. Thus, A ]. 9. D. sec cos cos cos sin ] cos cos. C. First, we must find the area of the region R: A 3 The -coordinate of the centroid is c b y A a 3 The y-coordinate of the centroid is y c b A y a 3 ) 3 Thus, the centroid is, ) 3. A. Using the Theorem of Pappus, the volume V ra, in which A is the area of the region and r is the distance from the centroid of the region to the line about which it is being rotated. Both A and the centroid were determined in the previous question, so all we have to find is the distance r. The centroid is, 3) and the line is + 3y 5. Using the formula for the distance from a point to a line r So V

4 6 FAMAT Convention Mu Integration ) sec sin ln. B. A + B + cos. Some careful observation would show that this integrand is just equal to d sec ln ) as found by the product rule. [ ] So we have sec ln ) ln a, b, c a + b + c 3. E. The limits tell us that we are integrating the region y, y. We know that we will be reversing the order of and y. The function that forms the lower boundary for y is y and the function that forms the upper boundary for y is y. The minimum value for is and the maimum value for is. So the reordered integral is e 3 dy.. A. After changing the order of integration in the last problem, this should be straightforward. [ ] y e 3 dy e 3 y e 3 y 3 e3] e 3 e sin 5. B. Using the hint provided, we set I) I ) e ) sin e sin. e sin e cos e cos e cos e sin + + ) e sin e cos e sin ) e sin e sin e ] + cos + sin ) + I ) d We still need to find I), so we find the antiderivative: arctan + C. We + are looking for a specific value of I), so we must find the particular solution and determine the value of C. Observation shows that lim I). So lim arctan + C + C C I) arctan. In this situation,, so I) arctan)

5 6 FAMAT Convention Mu Integration 5 6. C. L{cost)} ) + s e st cost) dt e st sint) + se st sint) dt e st sint) se st cost) s e st cost) dt e st sint) se st cost) e st cost) dt e st ] t + s sint) s cost)) t s + s e st cost) dt 7. B. sinh e e arcsinh ln + ) + arcsinh arcsinh + darcsinh ) + arcsinh + + C ln + ) C 8. C. Each unit step, the graph forms a right triangle with the -ais, each having an area of. So there are 3 full triangles above the -ais and below the -ais, plus smaller triangles. Thus, the integral is equal to 3 + 3) e ) 3 e + e + 3 A + B + C + D + E y {} y E. cot sec sec du tan ln u + C ln tan + C u 3. C. Because y has no established dependence on in this contet, y can simply be treated as a constant. Thus, y y] y