Calculus Math 21B, Winter 2009 Final Exam: Solutions

Transcription

1 Calculus Math B, Winter 9 Final Exam: Solutions. (a) Express the area of the region enclosed between the x-axis and the curve y = x 4 x for x as a definite integral. (b) Find the area by evaluating the definite integral in (a). Solution. (a) The area A is given by A = x 4 x dx. (b) Using the substitution u = 4 x, with du = x dx, we get A = = = 3 43/ = u du [ 3 u3/ ] 4

2 . Evaluate the following indefinite integrals: (a) x 4 x dx; (b) x tan x dx; (c) (9 x ) 3/ dx. Solution. (a) We use the substitution u = 4 x, with du = dx and x = 4 u. This gives x 4 x dx = (4 u) u du = (4u / u 3/) du = 8 3 u3/ + 5 u5/ + C = 8 3 (4 x)3/ + 5 (4 x)5/ + C. (b) We use integration by parts to differentiate the inverse tangent. Setting u = tan x, dv = x dx, du = + x dx, v = x, we get x tan x dx = x tan x x + x dx = x tan x ( ) dx + x = x tan x x + tan x + C. (c) We use the trigonometric substitution x = 3 sin θ (since the integral involves a square root of the form 3 x ). This gives dx = (9 x 3/ ( ) 9 9 sin θ ) 3 cos θ dθ 3/ = cos θ ( 9 sin θ ) dθ 3/

3 Using the identity sin θ = cos θ to simplify the integrand, we get (9 x ) dx = 3/ 9 cos θ dθ = sec θ dθ 9 = tan θ + C. 9 We have tan θ = sin θ cos θ = sin θ sin θ = = 3 sin θ 9 9 sin θ x 9 x. Thus, (9 x ) dx = x 3/ 9 + C. 9 x 3

4 3. For a >, let R be the region of the (x, y)-plane with x a that lies above the x-axis and below the curve (a) Find the area of R. y = x. (b) Let S be the solid formed by rotating R about the x-axis. volume of S. Find the (c) How do the area of R and the volume of S behave as a? Comment briefly on your answer. Solution. (a) The area A(a) of R is given by A(a) = a x dx = [ln x] a = ln a. (b) Splitting R in to vertical strips of height y and width dx, which give discs of radius y and thickness dx when rotated about the x-axis, we find that the volume V (a) of S is given by V (a) = a a πy dx = π x dx [ = π ] a x ( = π a ). (c) As a, we see that A(a) diverges to, whereas V (a) π converges to a finite limit. It may seem a bit odd that one can rotate an infinite area about an axis and get a finite volume; this is possible because areas closer to axis of rotation contribute less to the volume. 4

5 4. Consider the curve with parametric equation x = t sin t, y = cos t for t π. (a) Write down a definite integral for the length of the curve. (Don t evaluate the integral.) (b) Write down a definite integral for the area of the surface formed by rotating the curve about the x-axis. (Don t evaluate the integral.) Solution. (a) The arclength L is L = = = = = π π π π ds (dx ) ( ) dy + dt dt dt ( cos t) + sin t dt cos t + cos t + sin t dt cos t dt (b) The area A of the surface, with radius of revolution about the x-axis equal to y, is A = πy ds = π π = π ( cos t) cos t dt π ( cos t) 3/ dt. 5

6 Remark. This parametric curve is one loop of the cycloid shown on the next page. We can in fact evaluate these integrals. Use of the double angle formula ( ) t cos t = sin gives: π L = = 4 = 8; π A = 4 π = 6 π = 6 π = 3 π = 64π 3. cos t dt ( ) t sin dt π π π ( cos t) 3/ dt ( ) t sin 3 dt [ ( )] t cos sin ( u ) du ( ) t dt 6

7 π π π π 3π 4π Figure : A cycloid. 7

8 5. Define the improper integral x 3 e x dx as a limit of Riemann integrals, and evaluate it. Solution. The improper integral is defined by x 3 e x dx = lim b b x 3 e x dx. The Riemann integral on the right hand side exists because x 3 e x continuous on the interval x b for any b. is To evaluate the integral, we first use the substitution t = x, with dt = x dx, which gives where c = b. b x 3 e x dx = = b c Next, we use integration by parts with which gives c u = t, x e x x dx te t dt dv = e t dt, du = dt, v = e t, te t dt = [ te t] c c + e t dt = ce c + [ e t] c = e c ce c 8

9 It follows that x 3 e x dx = lim b b = lim c =. x 3 e x dx ( e c ce c) Here, we use the fact that ce c as c. This follows, for example, from l Hôspital s rule: ( lim ) ce c c = lim c c e = lim c c e =. c 9

10 6. Consider the Riemann integral + x dx, 4 and partition the interval x into n equal subintervals. (a) Write down the Riemann sum for the integral obtained by evaluating the integrand at the left endpoint of each subinterval. Does this Riemann sum provide a lower bound or an upper bound of the integral, or neither? Why? (b) Write down the Riemann sum for the integral obtained by evaluating the integrand at the right endpoint of each subinterval. Does this Riemann sum provide a lower bound or an upper bound of the integral, or neither? Why? Solution. The length of each subinterval in the partition is x = /n. We denote the endpoints of the intervals in the partition by {x, x, x,..., x n } where x k = k for k =,,,..., n. n A Riemann sum for the integral has the form n f(c k ) x = n k= n + c 4 k k= where x k c k x k. (a) If we choose c k = x k to be the left endpoint of the subinterval, the corresponding Riemann sum I n is given by I n = n n k= = n 3 n k= + [(k )/n] 4 n 4 + (k ) 4. The function /( + x 4 ) is decreasing on x, so its value at the left endpoint of a subinterval is larger than in the rest of the subinterval. Therefore this Riemann sum provides an upper bound for integral.

11 (b) If we choose c k = x k to be the right endpoint of the subinterval, the corresponding Riemann sum J n is given by J n = n n k= = n 3 n k= + (k/n) 4 n 4 + k 4. Since the function is decreasing, its value at the right endpoint of a subinterval is smaller than in the rest of the subinterval. Therefore this Riemann sum provides a lower bound for the integral.

12 7. Suppose that an empty reservoir has the shape of a circular cone of radius a at its surface and depth h, and the surface is at a height h above sea level. Calculate the work done against gravity required to fill the reservoir by pumping water from sea level into the reservoir. Express your answer in terms of a, h, and the weight density w = ρg of water. Solution. Let z be the height measure from above the bottom of the reservoir. (It would be equally correct to use height y = z + h above sea level instead.) The reservoir cross section at height z is a circle of radius r = az/h (since the radius varies linearly from r = at z = to r = a at z = h). The volume dv of a slice of the reservoir of thickness dz at height z is A(z) dz where A = πr is the cross-sectional area. The mass dm of water required required to fill this slice is ρa(z) dz, where ρ is the mass-density of water. Water at height z has to be pumped a vertical distance (z + h) from sea level. The work dw required to pump a mass dm of water is g(z + h) dm, or dw = ρg(z + h)a(z) dz = w(z + h)a(z) dz. Thus, the total work W required is W = = h h = πwa h = πwa h w(z + h)a(z) dz w(z + h)π h = 7 πwa h. ( az ) dz h ( z 3 + hz ) dz [ 4 z4 + 3 hz3 ] h

13 Figure : The Ben Cruachan reservoir. Remark. Although this problem is obviously artificial, there are technological uses of pumping water into reservoir. For example, the Ben Cruachan power station in Scotland is a hydroelectric power storage facility. During periods of low power demand, water is pumped a vertical distance of approximately 36 m from sea level to a reservoir in the mountain. During periods of peak demand, the water is released and its gravitational potential energy is converted back into electrical power by turbines located in a cavern excavated in the bottom of the mountain. 3

14 8. Define two functions f(x), g(x) by f(x) = x ln e t t dt, g(x) = x ln t dt. (a) For what x-values does the Riemann integral defining f(x) exist? Why? (b) For what x-values does the Riemann integral defining g(x) exist? Why? (c) If x >, show that f (ln x) = g(x). Solution. (a) The Riemann integral exists for any x >, since ln > and e t /t is continuous on any interval that does not contain t =. Since e t /t is discontinuous and unbounded at t =, the integral is not defined for x. (b) The Riemann integral exists for any x >, since / ln t is continuous on any interval in < t <. Since ln =, the function / ln t is discontinuous and unbounded at t =, so the integral is not defined for x. (c) Replacing x by ln x in the definition of f(x) we get f (ln x) = ln x ln e t t dt. Note that ln x > if x >, so f(ln x) is well-defined. Making the substitution t = ln u in this integral, with dt = u du, observing that u = when t = ln and u = x when t = ln x, and simplifying the result, we get f (ln x) = = = x x x = g(x). 4 e ln u ln u u ln u ln u du u du u du

15 Remark. The function f(x) is called an exponential integral, while g(x) is called a logarithmic integral, usually denoted by Li(x) = x ln t dt. Neither integral can be expressed in terms of elementary functions. The logarithmic integral is important in number theory. If π(x) denotes the number of primes less than or equal to x (for example, π(8) = 4 since, 3, 5, 7 are prime numbers), then π(x) is approximately equal to Li(x) for large values of x. This means that the density of prime numbers among the natural numbers near x is approximately equal to (/ ln x) when x is large. How π(x) fluctuates about Li(x) is not well understood, and is related to some of the outstanding unsolved problems in mathematics, such as the Riemann hypothesis. 5

16 9. Suppose that a function g(y) is given for < y < by g(y) = y t t dt. (a) Use partial fractions to evaluate the integral and find g(y). (b) Find the inverse function f of g by solving the equation x = g(y) for y = f(x). (You can assume that < y <.) Solution. (a) We use a partial fractions expansion of the form t t = t( t) = A t + B t. Putting the right-hand side of this equation over a common denominator, we get A t + B (B A)t + A =. t t( t) We therefore choose B A =, A =. The solution is A = B = /, meaning that t( t) = ( t + t ) It follows that g(y) = y ( t + ) dt t = [ln t ln t ]y = (ln y ln y ) = ln y y. 6

17 (b) For < y <, we have y/( y) >, so we may omit the absolute values in the logarithm, and write g(y) = ( ) y ln. y If x = g(y), then e g(y) = e x, or y y = ex. Multiplying this equation by ( y) and collecting the terms proportional to y, we get ( + e x ) y = e x. Thus, y = f(x) where f(x) = ex + e x. Remark. This integral arises in solving the logistic equation dy dx = y y. This nonlinear ordinary differential equation (ODE) describes the evolution of a population y(x) as a function of time x whose growth rate dy/dx decreases with increasing population y (e.g. due to the effects of overcrowding or limited resources). There are two constant solutions of this ODE, y = and y =. The solution y = corresponds to extinction, and y = corresponds to the maximum sustainable population. We can find general solutions of the ODE by the method of separation of variables. Formally, we write dy y y = dx and integrate to get y y dy = dx. 7

18 This integral gives If y(x) satisfies the initial condition then C =, and it follows that x = g(y) + C. y(x) = y() = ex + e x. Thus, the population grows from y = at x =, but the growth rate slows as y(x) increases, and y(x) approaches the maximum sustainable population as x. 8

19 . As a simplified model of the earth s atmosphere, consider a stationary atmosphere in which the pressure p(z) and mass density ρ(z) vary with height z <. Let g be the gravitational acceleration (assumed constant). (a) Explain briefly why p(z) = g z ρ(s) ds. (You can assume that the improper integral converges at infinity.) (b) Explain briefly why it follows that dp dz = gρ(z). (c) Suppose that ρ(z) = cp(z) for some constant c, meaning that the density is proportional to the pressure, and p() = p where p is the atmospheric pressure at the ground z =. Find p(z). (d) At what height H is the pressure p(h) = p /e reduced by a factor e from its value at the ground? Solution. (a) In hydrostatic equilibrium, the pressure at some height in a fluid is equal to the gravitational force per unit horizontal area exerted by the fluid above. This force is equal to g times the mass per unit area of the fluid above, or g z ρ(s) ds. (b) This equation follows from the fundamental theorem of calculus, and the fact that the integral changes sign when we exchange the limits: d dz z ρ(s) ds = d dz z ρ(s) ds = ρ(z). (c) Using the formula ρ = cp to eliminate the density ρ from the equation in (b), and the condition at z =, we get dp dz = cgp, p() = p. 9

20 The solution of this initial value problem is the exponential function p(z) = p e cgz. (d) The pressure decreases by a factor of e over a height H = cg. Remark. The pressure p, specific volume V, and absolute temperature T of a unit mass of an ideal gas are related by the equation pv = RT, where R is a gas constant. (For dry air, R = 87 JKg K, where the absolute temperature T is measured in Kelvins K.) The density of the gas is given by ρ = /V. Thus, ρ is proportional to p if the temperature T is constant, and in that case the constant of proportionality is given by c = RT. In an isothermal atmosphere in hydrostatic equilibrium, the density and pressure decrease exponentially with height, p(z) = p e z/h, ρ(z) = ρ e z/h. The height H over which they decrease by a factor of e is called the scale height of the atmosphere, and it is given by H = RT g. Although the earth s atmosphere is not exactly isothermal, its temperature in the first 7 km above the surface is approximately within 5% of 5 K. The corresponding scale height is H = 7. km. Humans require an air pressure of at least.5 atmospheres to survive on a long-term basis. This sets an upper limit on habitable altitudes of about 5.8 km or 9, ft e.g. for some communities in the Andes or Tibet. The peak of Mt. Everest, at an altitude of 8.85 km or 9, 8 ft, is not habitable.