Math 4, Final Exam, Fall 6, Solutions There are problems. Each problem is worth points. SHOW your wor. Mae your wor be coherent and clear. Write in complete sentences whenever this is possible. CIRCLE your answer. CHECK your answer whenever possible. No Calculators. If I now your e-mail address, I will e-mail your grade to you. If I don t already now your e-mail address and you want me to now it, then send me an e-mail. I will post the solutions on my website a few hours after the exam is finished.. Find sin x cos x + dx. Chec your answer. Let u = cos x +. Then du = sin x dx and the integral is equal to u / du = u + c = cos x + + C. Chec: The derivative of the proposed answer is /)cos x + ) / sin x).. Find sin 4 x cos 3 x dx. Chec your answer. The integral is equal to sin 4 x sin x) cos x dx. Let u = sin x. It follows that du = cos x, and the integral equals u 4 u ) du = u 4 u 6 ) du = u / u 7 /7 + C = sin x sin7 x 7 + C. Chec: The derivative of the proposed answer is sin 4 x cos x sin 6 x cos x = sin 4 x cos x sin x).
3. Find x x dx. Chec your answer. 4x + 8 The denominator does not factor. Complete the square: x 4x + 8 = x 4x + 4 + 4 = x ) + 4. Let u = x. It follows that du = dx and x = u +. The integral is equal to u + u + 4 du = u u + 4 + 4[u/) + ] du = / lnu + 4) + 4 4[w + ] dw = /) lnu + 4) + arctan w + C = /) lnu + 4) + arctanu/) + C = /) lnx ) + 4) + arctanx )/) + C, where w = u/ and dw = du/. Chec: The derivative of the proposed answer is x x 4x + 8 + ) = + x ) x x 4x + 8 + 4 ) x ) + 4 4. Find arctan x dx. Chec your answer. Use integration by parts. Let u = arctan x and dv = dx. Compute du = dx/ + x ) and v = x. The original integral is equal to xdx uv vdu = x arctan x + x = x arctan x /) ln + x ) + C. Chec: The derivative of the proposed answer is. Find Set x + x + arctan x x + x. x + 4 x 3 dx. Chec your answer. x x + 4 x 3 x = A x + B x + C x and solve for A, B, and C. Clear the denominator x + 4 = Axx ) + Bx ) + Cx.
Plug in zero to see that 4 = B ; so, B =. Plug in to see that 8 = 4C ; so, C =. Plug in to see that 6 = A ) + ) ; so, A =. Lets chec what we have so far: We see that 3 x + x + x = xx ) x ) + x x x ) = x + 4x x + 4 + x x x ) So the original integral is = x + 4 x 3 x. x + x + ) dx = ln x + + ln x + C. x x ) x x 6. Find lim. x x + This limit is equal to lim x Let t = x +. The limit is x + x + ) x + ) x. x x + lim + ) t + ) t + ). t t t t t I now that ) lim + r t t t = e r ; so, the answer to our question is e ) = e. 7. Find the area between y = x and y = x. I drew a picture elsewhere. The graphs intersect at 4, ) and, ). I chop the y -axis from y = to y =. I integrate the big x minus the little x. The area is y + y )dy = y / + y y 3 /3 = + 4 8/3 / + /3).
4 8. Consider the sequence {a n } with a =, and a n = + a n for n. Prove that the sequence {a n } converges. Find the limit of the sequence {a n }. Notice that a n for all n. It is clear that a <. If a n, then a n + ; hence, a n = a n + =. Our assertion is established by Mathematical Induction. We now claim that the sequence {a n } is an increasing sequence. We now that a n. Multiply both sides by the positive number a n + 4 to see that a n + 4a n a n +. In other words, a n a n +. The square root function is an increasing function; so, a n a n + = a n. Our claim is established. The sequence {a n } is an increasing sequence which never gets beyond. The Completeness axiom guarantees that the sequence converges. Let L be the name of lim a n. Tae the limit of a n = + a n to see that L = L +. We n can now solve for L. We have L = L + or L L =. We factor to get L )L + 4) =. So, L = 4 or L =. All of our a n are positive so L can not possibly negative. We conclude that L =.. A conical water tan sits with its base on the ground. The radius of the base is feet. The height of the tan is 3 feet. The tan is filled to a depth of feet. How much wor is required to pump all of the water out through a hole in the top of the tan? The density of water is 6.4 lb/ft 3. Be sure to give the units for your answer. I drew a picture elsewhere. Notice that I arranged my axis, so that x = is the top of the tan. The water starts at x =. The bottom of the water occurs at x = 3. For each x between and 3, we lift a thin layer of water starting at x -coordinate x. The wor to lift this thin layer is the weight of the layer times the distance this layer must be lifted. The distance is x. That is the advantage of the way I set my axis.) The weight of the layer is the volume of the layer times the density of water. The volume of the layer is the area of the top times the thicness. The thicness is dx and the area of the top is πr, where similar triangles tell us that r = 3x. The wor to lift the layer of water at x -coordinate x is The total wor is 6.4)π 3 x 3 dx = 6.4)π 6.4)π 3 x) xdx. x 4 4 3 = 6.4)π [3 4 4 ] foot-pounds. 36
. Consider the region in the first quadrant which is bounded by y = x, the x -axis, and x =. Revolve this region about the line y =. What is the volume of the resulting solid? I drew a picture elsewhere. I chop the y-axis from y = to y =. For each little piece of the y -axis I draw a rectangle from x = y to x =. I spin the rectangle and get a shell of volume πrht, where t = dy, h = y, and r = y. The volume of the solid is π y) y)dy = π ) = π /)y / /3)y 3/ y / + y y 3/ y y + )dy. Find the length of y = x 3/ from, ) to, ). The length is equal to ) dx + = dy dx = 4 + 3 x/) dx = [ 3 + 4 ))3/ + 4 )3/]. = π/) /3) / + ) + 4 xdx = 4 3 + 4 x)3/ x 3). Let fx) = 6. Find all real numbers x for which fx) = converges. Justify your answer. We use the ratio test. Let a + ρ a x 3 + 6 + + ) 6 x 3 x 3 6 + x 3 =. 6 If ρ <, then the series converges. If < ρ, then the series diverges. We see that ρ < precisely when 3 < x <. We also see that < ρ precisely when x < 3 or < x. We need only worry about x = 3 and x =. We see that 3) f) = 6 =, = = which is the harmonic series and diverges. We see that 6) f 3) = 6 = ), = which is minus the alternating harmonic series and converges. We conclude that = fx) converges for 3 x < and fx) diverges for all other x.
6 e x x x4 3. Find lim x6 x x 8 We now that It follows that and therefore, x. Justify your answer. e x = + x + x + x3 + x4 4! +... e x = + x + x4 + x6 + x8 4! +... ; e x x x4 lim x6 x x 8 ) + x + x4 + x6 + x8 4! +... x x4 x6 x8 x x x 8 ) x 8 4! +... x 8 ) 4! + x! + x4 6! +... x 8 ) x 4! + x! + x4 6! +... = 4. 4. Does = + converge? Justify your answer. Apply the limit comparison test with the divergent harmonic series that a lim b + = + + =, =. We see and is a number, not, not. We conclude that both series converge or both series diverge. The series diverges; hence, the series + also DIVERGES. =
. Does Notice that = +! + 3 converge? Justify your answer. +! + 3 < +! =! because the fraction on the right has a larger numerator and a smaller denominator. The series! converges by the ratio test: Both series = a + lim a = + +)!! +!+3 and smaller than the terms of = = + + )! Comparison test to conclude that the series 6. What is the exact sum of the series We now that if < x <, then x < x <, then ln x) =! + = <.! are positive series. The terms of!. The series = l= x l+ l+ = = +!+3 = +!+3 7 are! converges. We apply the also converges.? Justify your answer. 4 ) = x l. l= to see that = + C, so C = and ln x) = Integrate to learn that if +C, for some constant C. Plug in x = l= x l+ l+, for < x <. Let = l +. Notice that when l is, then =. Thus, if < x <, then ln x) = x. Plug in x = 4 to see that = = 4 ) = ln 4 ) = ln 3 4 ) = ln 4 3 ). 7. Approximate We see that = with an error at most. Justify your answer. n = = = =n+.
8 I drew some boxes elsewhere to help approximate the right most sum. The sum is the area inside the boxes, which is less than the area under the curve, which equals n We have shown that dx x b = 4x 4 b n n b 4b 4 + 4n 4 = 4n 4. = 4n 4 Notice that when n = 4 or higher) 4n < 4. We conclude that 4 = approximates = with an error at most. 8. Approximate answer. The given integral is equal to sinx ) dx with an error at most. Justify your x x6 + x! x4 7! +... dx = ) x 3 3 x7 7 + x! x 7! +... = 3 3 7 7 +! 7! +... We have found a series which converges to sinx ) dx. We may apply the alternating series test to the series. The series alternates. The absolute value of the) terms decrease. The terms go to zero. The distance between the sum of the entire series and some particular partial sum is less than the next term in the series. We see that 7 < 7. We conclude that 3 approximates the value of 3 sinx ) dx with an error at most.
. Find the Taylor polynomial P 3 x) of order 3 for the function fx) = lnx + ) about a =. We see that fx) = lnx + ), f x) = x +, f x) = x + ), f x) = f 4) x) = 6 x + ) 4, f) =, f ) =, f ) =, and f ) =. We now that P 3 x) = f) + f )x + f ) x + f ) x 3. Thus, x + ) 3, P 3 x) = x x + x3 3.. Keep the notation of problem. Find an upper bound for the error that is introduced if P 3 x) is used to approximate fx) when x <. We now that fx) P 3 x) = R 3 x) = f 4) c)x 4 4! = 6x 4 c + ) 4 4! = for some c between x and. We are told that x <.. So R 3 x) c + 4 4) 4. x 4 c + 4 4, We now that. < x <. and c is between x and. So,. < c <. and. < + c < +.. In other words,. < c +. It follows that c+ <, and R 3 x) 4 4 4) 4 = 4 4. We conclude that: if P 3 x) is used to approximate fx) when x <., then the error that is introduced is less than 4 4.