Examples. 1. (Solution) (a) Suppose f is an increasing function, and let A(x) = x

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1 Math 31A Final Exam Practice Problems Austin Christian December 1, 15 Here are some practice problems for the final. You ll notice that these problems all come from material since the last exam. You are, of course, still responsible for older material. As always, the questions and solutions are listed separately so that you can work each problem, and then click on (Solution) to see the solution. This document is mostly unedited, and almost certainly contains typos or other errors. If you spot any, please let me know. Finally, understand that this is just a sampling of problems that you should know how to do, and should not be considered an indication of the problems you ll see on the exam. Examples 1. (Solution) (a) Suppose f is an increasing function, and let A(x) = x f(t)dt. Is A necessarily an increasing function? Why or why not? (b) Can there exist a function f which is always increasing but whose area function A(x) = x f(t)dt is always decreasing? Why or why not?. Suppose f(v) gives the fuel efficiency (mpg) of your vehicle as a function of your vehicle s velocity, and suppose that v(t) gives your velocity (mph) at time t (measured in hours) on a road trip that lasts T hours. Set up an integral that represents the total fuel used (in gallons) for the trip. (Solution) 3. (Solution) (a) Compute (b) Compute [ ] d x f(t 5 )dt. dx x 3 [( ) d x 5 ( ) ] f(t)dt f(t)dt dx x 3. (Solution) (a) Find a function f(x) and a value x so that f has a root within 1 unit of x, but so that Newton s method will not terminate if we use an initial guess of x. (b) Suppose a function f has zeros at a, b, and c (and perhaps other zeros as well). If a < x < b, is it possible for Newton s method, with initial guess x, to converge to c? If so, give an example of such a function. 1

2 5. Suppose we know the following facts about the functions g and h: (a) g(x) > on (, 1/) and (3/, ), g(x) < on (1/, 3/), and g(1/) = g(3/) =. (b) h(x) > on (, 1), h(x) < on (1, ), and h() = h(1) = h() =. (c) g (x)h(x) + g(x)h (x) = g(x). (d) g(x + ) = g(x) (e) 1/ g(x)h(x)dx =. (f) c+1 c g(x)h(x)dx = for all c R. Use this information to sketch the graph of f on the interval [, ], where (Solution) f(x) = x g(t)h(t)dt. 6. A rocket is fired straight up, burning fuel at the constant rate of b kilograms per second. Let v = v(t) be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M = M(t) be the mass of the rocket at time t and note that M decreases as the fuel burns. If we neglect air resistance, it follows from Newton s Second Law that 1 F = M dv dt ub, where the force F = Mg. Thus, M dv dt ub = Mg. (1) Let M 1 be the mass of the rocket without fuel, M the initial mass of the fuel, and M = M 1 +M. Then, until the fuel runs out at time t = M b, the mass is M = M bt. You will need the fact that 1 dx = ln x + C. x (a) Substitute M = M bt into Equation 1 and solve the resulting equation for v. Use the initial condition v() = to evaluate the constant. (b) Determine the velocity of the rocket at time t = M /b. This is called the burnout velocity. (c) Determine the height of the rocket y = y(t) at any time t. Write the height as an integral, but do not solve this integral. 1 Found in Stewart s Essential Calculus. This problem uses the antiderivative of x 1, which wasn t covered this quarter, but you should be able to make the relevant substitution and use the given antiderivative to work your way through this one. Admittedly, it s a challenge problem.

3 (Solution) 7. Find the area of the ellipse bounded by x + y a b = 1. (Solution) 8. Let C be a circle centered at (c, ) with radius < r < c. By revolving C around the y-axis, we obtain a torus. Find the volume of this torus. (Solution) 9. The region enclosed between the curve y = kx and the line x = 1 k is revolved about the line x = 1 k. Use cylindrical shells to find the volume of the resulting solid, assuming k >. (Solution) 1. Evaluate the following definite and indefinite integrals : (a) (3 x) 1 dx (b) ax + bdx, where a (c) cos(3x) (d) π x sin(x )dx (e) (x + 3)(x 1) 5 dx (f) x x dx (g) (ax a 1 + b)(x a + bx) c dx, where a, b, c > 1. (h) a+π a (Solution) cos(x) sin(x)dx, where a is fixed. Some of these come from D.A. Kouba at UC Davis. 3

4 Solutions 1. (a) Suppose f is an increasing function, and let A(x) = x f(t)dt. Is A necessarily an increasing function? Why or why not? (Solution) No. By the fundamental theorem of calculus, A (x) = f(x), so A is increasing, but A is only increasing if A is always positive, which is not necessarily the case. In particular, suppose f(x) = x. Then A(x) = x /, which is not an increasing function. (b) Can there exist a function f which is always increasing but whose area function A(x) = x f(t)dt is always decreasing? Why or why not? (Solution) Yes. It s possible for a function to be always increasing, but never positive. Consider f(x) = 1/x on (, ). By graphing this, you should see that A is always decreasing, despite the fact that f is always increasing.. Suppose f(v) gives the fuel efficiency (mpg) of your vehicle as a function of your vehicle s velocity, and suppose that v(t) gives your velocity (mph) at time t (measured in hours) on a road trip that lasts T hours. Set up an integral that represents the total fuel used (in gallons) for the trip. (Solution) First notice that we can express instantaneous fuel efficiency as a function of time by f(v(t)). Our next observation is that the rate at which we re using fuel is given by dividing our speed by fuel efficiency. For example, if we re traveling 6 miles per hour in a vehicle which uses fuel at a constant rate of miles per gallon, then we re using fuel at a rate of 6/ = 3 gallons per hour. So if we let F (t) represent the amount of fuel we ve used after t hours of driving, then F (t) = v(t) f(v(t)), measured in gallons per hour. Then, according to the fundamental theorem of calculus, F (T ) = F (T ) F () = gives the total fuel used during the trip. 3. (a) Compute T [ ] d x f(t 5 )dt. dx x 3 v(t) f(v(t)) dt (Solution) Since the variable with respect to which we re differentiating appears in the bounds of our integral, this computation seems ripe for the fundamental theorem of calculus. But since these bounds aren t linear, we ll need to integrate

5 via substitution. Also, since the bounds are of different degrees, we ll need two substitutions, so we begin by splitting this into two integrals: [ ] [ d x f(t 5 )dt = d ] x f(t 5 )dt + f(t 5 )dt. dx dx x 3 x 3 For the first integral we can make the substitution u = 3 t, so that u 3 = t and 3u du = dt. We make this choice so that the x 3 in our lower bound will become x. Notice that our upper bound remains. For similar reasons, we ll use the substitution v = t for our second integral, so that v = t and v 3 dv = dt. Then d dx [ ] x f(t 5 )dt x 3 = d dx = d dx [ f((u 3 ) 5 )(3u du) + [ x x x ] 3f(u 15 )u du + d dx = 3x f(x 15 ) + x 3 f(x ), using the fundamental theorem of calculus. (b) Compute [( ) d x 5 ( ) ] f(t)dt f(t)dt dx x 3 ] f((v ) 5 )(v 3 dv) [ x ] f(v )v 3 dv (Solution) We immediately notice that this derivative will need the product rule: [( ) d x 5 ( ) ] [ ] f(t)dt f(t)dt = d x 5 ( ) f(t)dt f(t)dt dx x dx 3 x ( 3 ) x 5 [ d ] + f(t)dt f(t)dt. dx x 3 Next, we rewrite the two derivatives we re interested in so that they ll look like FTC-style derivatives. For the first, we make the substitution u 5 = t, so that 5u du = dt, and for the second we make the substitution v 3 = t, so that 3v dv = dt. Then d dx [( ) x 5 ( ) ] f(t)dt f(t)dt x 3 = d [ x ] ( ) ( ) x 5 [ f(u 5 )(5u d du) f(t)dt + f(t)dt dx x 3 dx x = d [ x ] ( ) ( ) x 5 [ 5u f(u 5 d x ] )du f(t)dt f(t)dt 3v f(v 3 )dv dx x 3 dx = 5x f(x 5 ) x 3 f(t)dt 3x f(x 3 ) 5 x 5 f(t)dt. ] f(v 3 )(3v dv)

6 Notice that our final answer still involves definite integrals, but this is okay.. (a) Find a function f(x) and a value x so that f has a root within 1 unit of x, but so that Newton s method will not terminate if we use an initial guess of x. (Solution) Consider the function f(x) = x 1/. This function has roots at x = 1/ and x = 1/, but if our initial guess is x =, then Newton s method will not be able to make even one improvement on this guess. Recall that the n th approximation using Newton s method is given by x n = x n 1 f(x n 1) f (x n 1 ). But f (x) = x, so f (x ) = and we cannot define x 1. You should be able to see that any time we encounter f (x n ) =, Newton s method will not be able to continue. Geometrically, this is because the tangent line to y = f(x) at (x n, f(x n )) is horizontal, and thus either does not intersect the x-axis or is the x-axis. In either case, we have no way of making a next approximation to the root. (b) Suppose a function f has zeros at a, b, and c (and perhaps other zeros as well). If a < x < b, is it possible for Newton s method, with initial guess x, to converge to c? If so, give an example of such a function. (Solution) Yes, this is possible. If f (x ) is very small (but nonzero) in absolute value, then x 1 can be very far from x, because the slope of the tangent line used for this approximation is very small and thus takes a long time to hit the x-axis. One example is given by the function f(x) = cos(x) and the initial guess x =.. This guess is straddled by the roots a = π/ and b = π/ of f, but Newton s method will make our next approximation x 1 = x f(x ) f (x ).69. Our function has another root at 3π/.71, and Newton s method will now cause our approximations to converge to this root rather than the roots which are closer to x. We see that even when Newton s method gives a root of a function, it does not always give the nearest root to our initial guess. This particular example can be seen below: 6

7 5. Suppose we know the following facts about the functions g and h: (a) g(x) > on (, 1/) and (3/, ), g(x) < on (1/, 3/), and g(1/) = g(3/) =. (b) h(x) > on (, 1), h(x) < on (1, ), and h() = h(1) = h() =. (c) g (x)h(x) + g(x)h (x) = g(x). (d) g(x + ) = g(x) (e) 1/ g(x)h(x)dx =. (f) c+1 c g(x)h(x)dx = for all c R. Use this information to sketch the graph of f on the interval [, ], where f(x) = x g(t)h(t)dt. (Solution) We start by finding the critical points of f. Notice that, by the fundamental theorem of calculus, f (x) = g(x)h(x), so f (x) = at x =, 1/, 1, 3/,. We have f() = ( ) 1 f = f(1) = ( ) 3 f = f() = 1/ 1 3/ = 1 g(t)h(t)dt = g(t)h(t)dt = g(t)h(t)dt = g(t)h(t)dt = 1/ g(t)h(t)dt + 1/ g(t)h(t)dt = 1 g(t)h(t)dt + g(t)h(t)dt =. 3/ 1/ g(t)h(t)dt Notice that f (x) > when g(x) and h(x) are both positive or both negative, and that f (x) < when g(x) and h(x) are nonzero and have opposite signs. So f is increasing on (, 1/), decreasing on (1/, 1), increasing on (1, 3/), and decreasing on (3/, ). We also see that f has a maximum value of, reached at 1/ and 3/, and that f has a minimum value of, reached at, 1, and. Finally, we check the concavity of f. We have f (x) = g (x)h(x) + g(x)h (x) = g(x). 7

8 Now we can recast our first fact to see that g(x) > on (, 1/) and (3/, 1), that g(x) < on (1/, 3/), and that g((1/)) = g((3/)) =. Because g(x+) = g(x), we also see that g(x) > on (1, 5/) and (7/, ), that g(x) < on (5/, 7/), and that g((5/)) = g((7/)) =. So f (x) > on (, 1/), (3/, 5/), and (7/, ), and f (x) < on (1/, 3/) and (5/, 7/). So f alternates concavity at 1/, 3/, 5/, and 7/. All of this gives us enough information to produce the following plot: 6. A rocket is fired straight up, burning fuel at the constant rate of b kilograms per second. Let v = v(t) be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M = M(t) be the mass of the rocket at time t and note that M decreases as the fuel burns. If we neglect air resistance, it follows from Newton s Second Law that F = M dv dt ub, where the force F = Mg. Thus, M dv dt ub = Mg. () Let M 1 be the mass of the rocket without fuel, M the initial mass of the fuel, and M = M 1 +M. Then, until the fuel runs out at time t = M b, the mass is M = M bt. You will need the fact that 1 dx = ln x + C. x (a) Substitute M = M bt into Equation and solve the resulting equation for v. Use the initial condition v() = to evaluate the constant. (b) Determine the velocity of the rocket at time t = M /b. This is called the burnout velocity. (c) Determine the height of the rocket y = y(t) at any time t. Write the height as an integral, but do not solve this integral. (Solution) 8

9 (a) When we make the substitution, we have (M bt) dv dt ub = (M bt)g dv dt = (bt M )g + ub. M bt So we can integrate the last equation to find v. Notice that t is the only variable quantity on the right hand side of this equation. We have dv (bt v(t) = dt dt = M )g + ub dt M bt ub = gdt + M bt dt. The first of these integrals is easy to evaluate; for the second, we make the substitution w = M bt, so we have dw = bdt. Then b 1 v(t) = gdt u M bt dt = gdt u w dw Since v() =, we have = gt u ln w + C = gt u ln M bt + C. = v() = u ln M + C, so C = u ln M. This means that v(t) = gt + u (ln M ln M bt ) = u ln M M bt gt. (b) We substitute t = M /b into our formula for v(t): ( ) M v = u ln M b M M g M = u ln b M M 1 g M b. (c) Since y (t) = v(t), the fundamental theorem of calculus says that t t t ( ) y(t) = y(t) y() = y (s)ds = v(s)ds = u ln M M bs gs ds. 7. Find the area of the ellipse bounded by x + y = 1. a b (Solution) It s safe to assume that a, b >, so we do. Also, because of the symmetry this ellipse has about both coordinate axes, we may calculate the area of the ellipse that lies in the first quadrant, and then multiply this area by. In the first quadrant, both x and y are positive, so we may solve for y to find that y = b 1 x /a 9

10 on this interval. Then the area bounded by y = b 1 x /a and the coordinate axes is given by a so the area of the ellipse is given by a b 1 x /a dx, b 1 x /a dx. We compute this by making the substitution x = a sin θ, so that dx = a cos θdθ. Notice that when x =, θ =, and that when x = a, θ = π/. So we have a b 1 x /a dx = = ab π/ π/ π/ b 1 sin θ(a cos θ)dθ π/ cos θ cos θdθ = ab cos θdθ = ab (1 + cos(θ))dθ = ab [θ + 1 ] π/ sin(θ) [( π ) ] = ab + ( + ) = πab. In particular, the area of a circle of radius r is πr, since in this case a = b = r. This serves as a bit of a reality check; if setting a = b = r had resulted in a formula for the area of a circle that doesn t agree with the formula we know, we would have had cause for concern. 8. Let C be a circle centered at (c, ) with radius < r < c. By revolving C around the y-axis, we obtain a torus. Find the volume of this torus. (Solution) You should draw a picture for this problem. We ll use the disk/washer method, so our integral will be with respect to y. For a fixed y value, we will be rotating a strip which is parallel to the x-axis around the y-axis; we need to determine the distance of each end of this strip to the y-axis. The smaller distance is called the inner radius, and the greater distance is called the outer radius. Now our circle is determined by the equation (x c) + y = r, so we may solve this to find that the endpoints of our horizontal strip are x = c + r y and x = c r y. That is, our inner and outer radii are given by R I (y) = c r y and R O (y) = c + r y. The area of the cross-section obtained by revolving this strip about the y-axis is then given by A = π(r O (y)) π(r I (y)), so the volume of the torus is given by integrating 1

11 this area from the smallest possible value of y to the largest 3. Here s a picture of what we re doing; the blue strip represents the horizontal strip we re revolving about the y-axis. We easily see that the smallest and largest values of y are r and r, so r V = π [ (R O (y)) (R I (y)) ] r dy = π [(c + r y ) (c ] r y ) dy = = r r r r r [ π r (c + c r y + (r y )) (c c ] r y + (r y )) dy πc r y dy = πc r r r y dy. This last integral will need a substitution. Whenever we have a term that looks like r y, it s a good idea to try the substitution y = r sin θ. This will lead us to r y = r r sin θ = r (1 sin θ) = r cos θ, which will help in this integral. Notice that using this substitution gives dy = r cos θdθ. Also, we have y = ±r when θ = ±π/, so our integral becomes V = πc = πc = πr c π/ π/ π/ = π r c. π/ r r sin θ(r cos θ)dθ = πc π/ r cos θdθ = πr c [ θ + sin(θ) ] π/ π/ π/ = πr c π/ π/ 1 + cos(θ) dθ [ (π ) + r cos θ(r cos θ)dθ ( )] π + Just for fun, here s a picture of a torus, rendered in Mathematica: 3 This is a very rapid explanation of our integral; you can also see the notes/textbook to determine why our integral is set up this way. 11

12 9. The region enclosed between the curve y = kx and the line x = 1 k is revolved about the line x = 1 k. Use cylindrical shells to find the volume of the resulting solid, assuming k >. (Solution) The region we re interested in revolving is pictured here, with a particular value of k = 3: In this plot, the shaded region is the region we will revolve, and we will revolve this region about the dashed, vertical line on the right, x = k. The blue line in our region represents a strip that we ll use in finding the volume of the resulting solid. For each one of these vertical strips say based at x the surface area of the cylinder that results from revolving this strip about the line x = k is πr(x)h(x), where r(x) is the radius of the resulting cylinder, and h(x) is the height of the strip. Since we re revolving about x = k, the radius should be given by r(x) = k x. By rewriting y = kx, we see that the height of our strip is h(x) = kx. 1

13 The volume of the resulting solid is then obtained by adding together the surface areas of all of these cylinders, as x runs from to k/. That is, k/ k/ ( ) k kxdx V = πr(x)h(x)dx = π x = π k/ k 3/ x 1/ dx π k/ k 1/ x 3/ dx [ ] k/ [ ] k/ = πk 3/ 3 x3/ πk 1/ 5 x5/ = πk 3/ 3 ( k ) 3/ πk 1/ 5 = π 1 k3 π k3 = 7πk3 1. ( k ) 5/ 1. Evaluate the following definite and indefinite integrals (a) (3 x) 1 dx (b) ax + b dx, where a (c) cos(3x) (d) π x sin(x )dx (e) (x + 3)(x 1) 5 dx (f) x x dx (g) (ax a 1 + b)(x a + bx) c dx, where a, b, c > 1. (h) a+π a (Solution) cos(x) sin(x)dx, where a is fixed. (a) Use the substitution u = 3 x, so that du = dx. Then our integral becomes (3 x) 1 dx = u 1 ( du) = u11 x)11 (x 3)11 + C = (3 + C = + C (b) Use the substitution u = ax + b, so that du = adx, meaning that dx = 1 du. So a our integral is ax u 1 + b dx = a du = 1 ( ) a 3 u3/ + C = 3a (ax + b)3/ + C. (c) This one is not difficult to do directly, but we ll use substitution anyways. Let u = 3x, so that du = 3dx. Then dx = 1 du, so 3 cos(3x) = cos(u) 1 3 du = 3 sin(u) + C = sin(3x) + C. 3 13

14 (d) Since we know an antiderivative for sin(x), but not for sin(x ), let s make the substitution x = u, so that xdx = du. When x =, u =, and when x = π, u = π, so our integral becomes π x sin(x )dx = 1 π = 1 [ cos(u)]π sin(x )(xdx) = 1 π = cos() cos(π) sin(u)du (e) This may not immediately seem like an integral that s well-suited for substitution, but making a clever substitution will allow us to avoid a lot of algebra. Let x = u + 1, so that dx = du, and in particular so that x 1 = u. Then our integral becomes (x + 3)(x 1) 5 dx = (u + )u 5 du = u 6 du + u 5 du = u u6 + C = (x 1) (x 1)6 + C. = 1. (f) Use the substitution u = x. So du = xdx. Our integral is then x x dx = 1 x ( x)dx = 1 udu = 1 3 u3/ + C = 1 3 ( x ) 3/ + C. (g) Use the substitution u = x a + bx, so that du = (ax a 1 + b)dx. Then our integral becomes (ax a 1 + b)(x a + bx) c dx = u c du = uc+1 c C = (xa + bx) c+1 + C. c + 1 (h) First we make the substitution u = t a, so that du = dt. Then, since u = when t = a and u = π when t = a + π, a+π a cos(t) sin(t)dt = π cos(u + a) sin(u + a)du. Next we make the substitution v = sin(u + a), so that dv = cos(u + a)du. Then when u =, v = sin(a) and when u = π, v = sin(π + a) = sin(a), so a+π a cos(t) sin(t)dt = sin(a) sin(a) [ v vdv = ] sin(a) sin(a) = ( sin(a)) (sin(a)) =. 1