MATH 101 Midterm Examination Spring 2009

Transcription

1 MATH Midterm Eamination Spring 9 Date: May 5, 9 Time: 7 minutes Surname: (Please, print!) Given name(s): Signature: Instructions. This is a closed book eam: No books, no notes, no calculators are allowed!. An eam booklet for your rough work has been provided; it will not be collected. 3. There are ten conceptual questions and ten long answer questions; for the long answer questions you must show all your work.

2 MATH Midterm Eamination Maimum Your Mark 3 Conceptual Long Answer Total

3 MATH Midterm Eamination 3 Conceptual Questions. Suppose that the function f() is continuous on the iterval [a, b]. Then according to the Fundamental Theorem of Calculus Part, the function G() defined for a b by G() = a f(t) dt satisfies G () = f(). Suppose that the function f() is continuous on the iterval [a, b]. Then according to the Fundamental Theorem of Calculus Part, if F is any antiderivative of f, then b f() d = F F a 3. Suppose that R is a plane region enclosed by the curves = f(y) and = g(y) and the lines y = c and y = d. Assuming that f(y) g(y) for all y [c, d], write a formula for obtaining the area of the region R. A = d[ ] c f(y) g(y) dy 4. Suppose the solid S is sliced by planes perpendicular to the y ais, and the area of each cross-section is A(y), where c y d. Write a formula for the volume of the solid S. V = d A(y) dy c 5. Suppose that R is a plane region enclosed by the curves y = f() and y = g() and the lines = a and = b, where < a < b and f() g() for all [a, b]. If S is the solid obtained by rotating the region R about the y-ais, write the formula for obtaining the area of the solid S using cylindrical shells. V = b a π[ f() g() ] d

4 MATH Midterm Eamination 4 6. Write the formula of integration by parts. u dv = uv v du or u()v () d = u()v() v()u () d 7. Write the trigonometric identities for the sine/cosine of the sum of two angles. sin( + y) = sin cos y + cos sin y cos( + y) = cos cos y sin sin y 8. Write the identities for the sine/cosine of the double angles. sin() = sin sin y cos() = cos sin 9. Write the half-angle identities for epressing sin or cos in terms of cos(). sin = cos() cos = +cos(). Fill in the blanks, about the convergence/divergence of a p-series. convergent, if p > ; is n p divergent, if p. n=

5 MATH Midterm Eamination 5. Find the derivative of the function g() = Long Answer Questions You must show all your work. Solution. This is a direct application of FTC: 5 dt y = + t 4 g () = d + 8 d ( ) = + 8. cos cos(u ) du. First we need to split the integral by additivity: y = 5 cos cos(u )du = 5 and now we use FTC to compute the derivative: cos(u )du cos cos(u )du, y = cos(5 ) d d (5) cos(cos ) d d (cos ) = 5 cos(5 ) + cos(cos ) sin.

6 MATH Midterm Eamination 6. Evaluate the indefinite integral dt cos t + tan t d. Solution. Use the substitution u = + tan t. Then du = sec tdt, and we have dt cos t + tan t = sec tdt + tan t du = = u / du = u / + c = ( + tan t) / + c. u Here we use u =, so du = d and = u. ( u) ( u + u ) d = ( du) = du u u = (u / u / + u 3/ )du = (u / 43 u3/ + 5 ) u5/ + c = ( )3/ 5 ( )5/ + c.

7 MATH Midterm Eamination 7 3. Find the area enclosed by the curves 4 + y = and = y. Find the volume of the resulting solid if the region under the curve y = +3+ from = to = is rotated about the y-ais. Solution. First find the intersection points: 4 + = + 4 = ( + 6)( ) = = 6, or. The corresponding y-values are y = 6, or. Now we compute the area enclosed using horizontal area elements (draw a picture!): A = 6 [ ( 4 y + 3) y ] dy = [ y3 y + 3y ] 6 = In this case we use cylindrical shells, so the volume is V = π To evaluate the integral we use partial fractions: d = π ( + )( + ) d. ( + )( + ) = A + + B + = (A + B) + A + B. Equating coefficients we get A + B = and A + B =, and solving this system A = and B =. Thus the volume is [ V = π + + ] d + = π [ ln + + ln + ] = π[ ln 3 3 ln ] = π ln 9 8.

8 MATH Midterm Eamination 8 4. Evaluate the integral ln 3 d π e cos t sin(t)dt. Solution. Here we use integration by parts: { use u = ln du = d dv = 3 d v = [ ln d = ln 3 [ = ln ] ] + + d 3 [ 4 ] = 3 6 (ln ). 8 First let s make the substitution = cos t, so that d = sin tdt. Then π e cos t sin(t)dt = = π e cos t sin t cos tdt = π e ( d) = e d { u = now by parts dv = e d ( ) = [e ] e d e cos t cos t(sin tdt) du = d v = e = ( e ( e ) [e ] ) = ( e + e [e e ] ) = 4e = 4 e.

9 MATH Midterm Eamination 9 5. Evaluate the integral du u 5 u d cos. Solution. Here we use u = 5 sin θ, so du = 5 cos θ dθ. Then du u 5 u = = 5 5 cos θ dθ 5 sin θ 5 cos θ csc θ dθ = ln csc θ cot θ + c 5 = 5 5 u ln 5 u u + c = 5 5 u ln 5 u + c. d cos = cos + = cos d = = cos cos + cos + d cos + sin d ( cot csc csc ) d = csc + cot + c.

10 MATH Midterm Eamination 6. Evaluate the indefinite integral d + /3 cos sin + sin d. Solution. To eliminate the fractional power 3 we use the substitution = u3, so then d = 3u du, and we have 3u du d = + /3 u 3 + u = 3 u du u + = 3 ln(u + ) + c = 3 ln(/3 + ) + c. We set u = sin, so du = cos d. Then cos sin + sin d = du u + u = u(u + ) du [ = u ] du u + = ln u ln u + + c = ln u u + + c = ln sin sin + + c.

11 MATH Midterm Eamination 7. Find the Midpoint Rule approimation M 5 for e/ d. [You don t need to compute the approimation as a decimal number, it is enough to stop once you have evaluated the appropriate function at the appropriate points.] How large do we have to choose n so that the approimation M n to the integral e/ d is accurate to within.? Solution. In this problem the integrand is f() = e /, the endpoints of the interval of integration are a = and b =, and the number of subintervals is n = 5, so = =. 5 5 Using the formulas for the Midpoint Rule we get M 5 = 5 [ f(.) + f(.3) + f(.5) + f(.7) + f(.9) ] = 5[ e /. + e /.3 + e /.5 + e /.7 + e /.9]. In order to estimate the errors we need f. f () = e/ f () = + 4 e /. In fact computing one more defivative you may see that f () is negative on [, ], this says that f () is decreasing on [, ], so f () f () = 3e. We are to guarantee accuracy to within. = 4. Taking K = 3e, using the error bound for the Midpoint Rule, E Mn 3e( ) 4 n 4 e 4n 8 n 4 e 8 = e e = , so to get the desired accuracy it suffices to take n an integer larger than 5 e/, say n = 59.

12 MATH Midterm Eamination 8. Determine whether the improper integral is convergent or divergent. arctan d ( + ) π sin d. Solution. First observe that for any real number, we have arctan π and +. Necessarily this is true for any [, ) (the interval of integration). Then arctan ( + ) d π = π π = π ( + ) d ( + ) d + π ( + ) d + π ( + ) d + π ( + ) d ( ) d 3 d In the last line the first integral is finite (and you can evaluate it using a substitution, verify this!), while the second integral is a convergent improper integral by the theorem on p-integrals (part ). Then by the comparison theorem, the given improper integral is convergent. Note that, since sin, we have sin, sin. Also, using the the theorem on p-integrals, part, we have π π d = π d = / = π. Then, by comparison, we conclude that π sin d is convergent.

13 MATH Midterm Eamination 3 9. Determine whether the series is convergent or divergent. n= n + 3 n n + n ( ) n= n+ e/n n. Solution. Observe the terms of this series: n+3 n n+ n, and notice that the dominating term in the numerator is 3 n, while the dominating term in the denominator is n. This means that for large values of n, the terms of the series behave like 3n n. Let s compare these terms in the limit: lim n n+3 n n+ n 3 n n = lim n n+3 n 3 n n+ n n = lim n n + 3 n n + = n Since the limit is a finite positive number, and since the series 3 n n= is divergent, by the Limit n Comparison Test we conclude that the given series is divergent. Since this is an alternating series, let s see if the conditions of the AST are satisfied. To see that the sequence b n = e/n is decreasing, let s look at the related function f() = e/ defined for n. Using the quotient rule for derivatives, we have f () = e/( ) e / = e/ e /. Since f () <, the function is decreasing, and therefore so is the sequence e/n. n To compute the limit of the sequence b n = e/n as n, we also look at the related function. n Since implies <, we see that e/ e, and using the Squeeze Theorem: e / lim = lim e = e Thus lim / e =, and therefore lim /n =. Now that we know both conditions of the n AST are satisfied, we conclude that the given series is convergent.

14 MATH Midterm Eamination 4. Determine whether the given series is absolutely convergent, conditionally convergent, or divergent. ( n) n ( ) 5n n. n! n + n= Solution. This is a tipical eample where we want to think of the Ratio Test. L = lim a n+ n a n [ ] ((n + )) n+ n! = lim n (n + )! (n) [ n ] (n + ) n (n + ) n! = lim n (n + )n! (n) ( ) n n n + = lim n n ( = lim + n = e. n n) Since L = e >, by the Ratio Test the series is divergent. n= Here we weant to think of the Root Test. We have L = lim 5n n n n a n = lim n n n + ( ) 5 n = lim = n n + ( ) 5 = 3. Since L = /3 <, by the Root Test the series is absolutely convergent.