Calculus I Sample Final exam

Transcription

1 Calculus I Sample Final exam Solutions [] Compute the following integrals: a) b) 4 x ln x) Substituting u = ln x, 4 x ln x) = ln 4 ln u du = u ln 4 ln = ln ln 4 Taking common denominator, using properties of the logarithm, one can get the answer in the more compact form sinθ)e sin θ) dθ 4 x ln x) = ln Substituting u = sin θ, du = sin θ cos θ dθ = sinθ) dθ, sinθ)e sin θ) dθ = e u du = e u + C = e sin θ + C c) d) x x + Substituting x = tan θ, x + = sec θ, = sec θ dθ, x x + = sec θ cos θ tan θ sec θ dθ = sin θ dθ = sin θ + C = x x + + C π x sin x Integrating by parts, π π π π x sin x = x cos x + x cos x = π + x cos x π π ) π = π + x sin x sin x = π sin x π = π + cos x = π 4

2 e) t + 8 t 5t + 6 dt Since the numerator has degree equal to that of the denominator, we have to divide to get t + 8 t 5t + 6 = + 5t + t 5t + 6 Next we decompose 5t + t 5t + 6 into partial fractions: 5t + t 5t + 6 = 5t + t )t 3) = A t + B t 3 Using Heaviside s method we see at once that A =, B = 7 so that, all in all, t + 8 t 5t + 6 = + 7 t 3 t Thus t + 8 t 5t + 6 dt = + 7 t 3 ) dt = t + 7 ln t 3 ln t + C t f) / x + x x + By partial Fractions x + x x + = x + x ) = A x + B Ax ) + B = x ) x ) Equating the numerators of the expression at the extreme right with that at the extreme left we get Ax + B A) = x +, thus A =, B A =, thus B = 4 Now / x + x x + = = / x + 4 ln x 4 x ) x ) ) / = 4 + ln [] Find the volume of the solid generated by revolving the region bounded by the following curves about the x-axis Draw a picture of the region to be revolved {y = x}, {y = }, {x = } The region in question is

3 3 By slicing washers V = π x) ) = π 4 4x) = π By the shell method Here we describe the region in terms of y; it is the region between the y axis and the curve x = y /4 for y V = π ) y y dy = π 4 y 3 dy = π [3] Find the center of mass of a thin plate of constant density δ covering the region delimited by {y = x }, {y = } The region in red is the region covered by the plate:

4 4 By symmetry, x = Now assume, as one may, that δ = ) M x = M = x ) ) = x ) = x 4 ) = 6, 5 Thus ȳ = M x /M = 6/5 The center of mass is at, 6 ) 5 [4] It took 8 J of work to stretch a spring from its natural length of m to a length of 5m Find the spring force constant 8 = W = k 3 x = 9k Solving for k, k = 4 [5] A tank in the shape of a circular cone with its vertex at the bottom, base at the top, is filled with a liquid that happens to weigh Newton/cubic meter The tank is meters high, the radius of the base of the cone is 4 meters Determine the works it takes to empty the tank if the liquid needs to be pumped to a height meters above the top of the tank Suggestion: Use the axis of the cone as the y-axis, with the origin at the base of the cone The top of the tank is then at y = and the water needs to be raised to y = 4 Consider a slab of liquid at level y of thickness dy By similar triangles this is an infinitesimal cylinder of radius y/3, height dy, thus of volume dv = πy/3) dy = πy dy/9 Since the liquid weighs Newton per cubic foot, dv equals the weight of the slab in Newtons The slab needs to be raised 4 y meters Thus W = π y)y dy = 3π [6] Determine the nature convergence or divergence) of the following integrals and give reasons in each case a) s + 4 s ds s + b ds = lim 4 s b = lim b s + ds = lim 4 s b 4 b ) / + + arcsin b 4 s ) / + arcsin s ) b ) = + π Converges

5 5 b) x x This integral can be computed But, since c) lim x x x x =, and /x ) converges being a p integral wit p = >, the integral converges x e x3 Converges One can compare to e x, or one can calculate it [7] Determine the following limits, or declare that the corresponding sequence diverges a) lim n e n + e n ) n n b) lim n n + c) lim n! n n e n + lim n e n = ) n n lim = e n n + n lim n n! =

6 6 [8] Determine the nature convergence or divergence) of the following series Give reasons in each case a) b) lnn) n Since lim n ln n n n = lim n ln n = the series diverges by limit comparison with the divergent harmonic series /n /n) lnn) ln n) n= One approach is to state that we saw in class that the series n= /[nln n)p ] converges if and only if p > not exactly a p series!), thus the series in question converges by limit comparison with the series n= /[nln n) ] Or one can apply the integral test directly; the terms of the series decrease: The series converges b x ln x ln x = lim b x ln x ln x ln b du = lim b ln u u = ln x) u ln b = lim arcsec u b = π arcsec ln ln c) ln n n) n n The series converges by the root test d) e) ) n n n n= The series diverges by the n-th term test n 6 + n + n 3 ) Converges Done in class [9] Compute 6 n )n + )

7 7 This is a telescoping series One has 6 k )k + ) = 3 k ) k + thus Thus n 6 s n = k )k + ) k= n = 3 k ) = 3 k ± + n ) n + k= = 3 ) 3 as n n + [] Determine which of the following series: i Converge absolutely ii Converge conditionally iii Diverge Justify your answer a) ) n+ n 3/ 6 n )n + ) = 3 It converges absolutely because )n+ n 3/ is a p-series with p = 3/ >, thus converges = n 3/ b) ) n+ n The series converges by the alternate series test However, the series of absolute values is a p-series with p = / <, hence diverges The series converges conditionally

8 8 [] Find the radii and intervals of convergence of the following power series a) b) c) x n+ n! x n 3 n n n The radius is r =, the interval is, ) The radius is r = 3, the interval is [ 3, 3] n= n n + x 3)n The radius is r =, the interval is [ 5, frac7 ) d) ) n x + ) n 3 n n= The radius is r = 3, the interval is 4, ) [3] Find the Taylor polynomials of order 3 generated by f at x = when a) fx) = ex + e x Method We know or have no excuse for not knowing) that e x = +x+ x + x3 + x4 + ; replacing x by x one gets 6 4 e x = x+ x x3 + x4 ; 6 4 adding and dividing by two, e x + e x )/ = + x + x4 + Keeping only the 4 terms of degree 3 we get that the polynomial is: + x Method We compute three derivatives: f x) = ex e x, f x) = ex + e x, f x) = ex e x Setting x = also in fx)) we get that f) =, f ) =, f ) =, f ) = The third order Taylor polynomials at x = is: b) fx) = sinx) T 3 x) = f) + f )x + f )x + 6 f )x 3 = + x

9 9 Method We know or have no excuse for not knowing) that sin x = x x3 + x5 ± Keeping only the terms of degree 3 we get that the 6 polynomial is: x x3 6 Method We compute three derivatives: f x) = cos x, f x) = sin x, f x) = cos x Setting x = also in fx)) we get that f) =, f ) =, f ) =, f ) = The third order Taylor polynomials at x = is: T 3 x) = f) + f )x + f )x + 6 f )x 3 = x x3 6 [4] Write out a Maclaurin series for the following functions fx) In each case give the interval of validity in what interval is fx) equal to the sum of its series) a) fx) = x e x Remember to state the interval where the Maclaurin series converges to fx) Since e x = x n n=, and this is valid for < x <, we get by n! the simple expedient of multiplying by x, x e x = n= x n+ n! = x + x 3 + x4 + x5 6 + The interval of validity is, of course, still < x < b) fx) = 4x Remember to state the interval where the Maclaurin series converges to fx) We saw zillions of time that x = x n if and only if) x < n= Thus replacing x by 4x) 4x = 4 n x n = + 4x + 6x + 64x 3 + n=

10 for 4x < Because if a power series centered at converges to a function in an interval around zero then the power series is the Maclaurin series, the series in the frame above is the Maclaurin series for / 4x) Because it converges to the function if and only if 4x <, the interval of validity is 4, 4 ) c) fx) = arctan x Done in class