WORKSHEET 1 SOLUTION Chapter 2 Differentiation

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1 United Arab Emirates University College of Sciences Department of Mathematical Sciences WORKSHEET SOLUTION Chapter Differentiation Calculus I for Engineering MATH SECTION CRN : :5 on Sunday & Tuesday Due Date: Tuesday, September, 8 ID No: Solution Name: Solution Score: / Instructions:. Read the questions carefully before you start working.. Show all your work to get the full credit.. Make your work to be neat and organized.. Do NOT use the calculator.

2 Calculus I for Engineering FIRST WORKSHEET SOLUTION Fall, 8. (Total 5 points) Find the derivative of each function. (Do NOT simplify your answer.) (.) (5 points) f () = + 5sec + ln Answer. We simplify first: f () = + 5sec + ln. Differentiating it, we get f () = + 5 ( sec )( tan ) () +, i.e., f () = + sec tan +. (.) (5 points) g(t) = et t + 5t Answer. g (t) = (et ) (t) e t (t) (t) + 5 t (ln5)(t ) = et (t) e t 6t + 5 t (ln5)(t) = et (t ) t + t5 t ln5. (.) (5 points) r(θ) = θ aθ cos (θ), where a is a constant. Answer. r (θ) = 6θ a [(θ) cos (θ) + θ ( cos (θ) ) ] = 6θ a [ cos (θ) + θ (cos(θ))( sin(θ))(θ) ] = 6θ a [ cos (θ) + θ (cos(θ))( sin(θ)) ] = 6θ acos(θ)[cos(θ) 8θ sin(θ)].. (5 points) Find a such that the tangent line to the graph of f () = a at = has the slope 6. Answer. The tangent line to the graph of f () at = has the slope f ( ). We compute f () = a, f ( ) = a. The condition f ( ) = 6 implies a = 6, i.e., a =. Page of Points Earned: out of points.

3 Calculus I for Engineering FIRST WORKSHEET SOLUTION Fall, 8. (Total points) A particle moves along a line so that at time t its position is given by s(t) = 6t t, where s(t) in meters and t in seconds. (.) ( points) What is its initial velocity v()? Answer. The velocity v(t) is obtained by v(t) = s (t) = 6 t. So the initial velocity is v() = 6 m/s. (.) ( points) When does it change the moving direction? Answer. The particle changes the direction when the velocity is zero. We observe v(t) = 6 t = at t =. Thus, it changes the direction at t =. (.) ( points) How fast is it moving when it returns to the initial position? Answer. The particle is located at the initial position when s(t) =. We observe s(t) = 6t t = t(6 t) = at t = or t = 6. It implies that the particle returns to the initial position at t = 6. Moreover, the velocity at t = 6 is v(6) = 6 (6) = 6 m/s. Hence, the desired speed is 6 m/s. (The negative sign means that the moving direction has been changed.) Page of Points Earned: out of points.

4 Calculus I for Engineering FIRST WORKSHEET SOLUTION Fall, 8. (Total 5 points) Let m A, m B, m C, m D and m E be the slopes of the tangent lines at the points A, B, C, D, and E on the given graph, respectively. (.) ( point) Among the slopes, m A to m E, which one is the smallest one? (.) ( point) Among the slopes, m A to m E, which one is the largest one? (.) ( points) List the slopes m A, m B, m C, m D and m E in increasing order from the smallest one. Answer. Let us compare the values m A, m B, m C, m D and m E. (i) It is easy to see that m A <, m B >, m C >, m D =, and m E <. That is, It gives four possible lists: m A, m E < m D = < m B, m C. m A < m E < m D < m B < m C, m A < m E < m D < m C < m B, m E < m A < m D < m B < m C, m E < m A < m D < m C < m B. (ii) It looks to me that m A < m E, which can be observed by drawing tangent lines at A and E. The tangent line at A is steeper than the one at E. So, the possible lists are m A < m E < m D < m B < m C, m A < m E < m D < m C < m B. (iii) The point D is closer to C than to B. It implies m C < m B. Therefore, the desired list is m A < m E < m D < m C < m B. 5. (5 points) Find the second degree polynomial (i.e., f () = a + b + c) satisfying all the conditions: f () =, f () = 5, f () =. Answer. From f () =, we get = f () = a( ) + b() + c = c, i.e., c =. From f () = 5, we get f () = a + b, 5 = f () = a() + b = b, i.e., b = 5. Page of Points Earned: out of points.

5 Calculus I for Engineering FIRST WORKSHEET SOLUTION Fall, 8 From f () =, we get f () = a, = f () = a, i.e., a =. Therefore, we deduce f () = + 5 = ( + ). Page of Points Earned: out of?? points.

6 United Arab Emirates University College of Sciences Department of Mathematical Sciences WORKSHEET SOLUTION Chapter Applications of Differentiation Calculus I for Engineering MATH SECTION CRN : :5 on Sunday & Tuesday Due Date: Sunday, October 9, 8 ID No: Solution Name: Solution Score: / Instructions:. Read the questions carefully before you start working.. Show all your work to get the full credit.. Make your work to be neat and organized.. Do NOT use the calculator.

7 Calculus I for Engineering SECOND WORKSHEET SOLUTION Fall, 8. (5 points) Find the absolute etrema of the function f () = Answer. on the interval (,). f () = = + = +, f () = ( ). Sign of f + Graph of f Increase f () = Decrease As approaches to from the right hand side of, the values of f () become negatively smaller and smaller, i.e., the graph of f () goes downward near = from the right hand side of. Similarly, as approaches to from the left hand side of, the values of f () become negatively smaller and smaller, i.e., the graph of f () goes downward near = from the left hand side of. It implies f () does not have an absolute minimum value on the interval (,). However, by the First Derivative Test and the table above, we deduce f () has the absolute maimum value f () =. 5 y Graph of f () = Page of 5 Points Earned: out of 5 points.

8 Calculus I for Engineering SECOND WORKSHEET SOLUTION Fall, 8. (5 points) Find the intervals where the graph of f () = +,, is increasing and decreasing. Answer. f () = + = +, f () = = ( )( + ) =. Sign of f + + Graph of f Increase f ( ) = Decrease f () = Increase We observe that f () > for < or > and f () < for < <. Hence, f () increases on (, ) (, ) and f () decreases on (,). 5 y Graph of f () = + = + Page of 5 Points Earned: out of 5 points.

9 Calculus I for Engineering SECOND WORKSHEET SOLUTION Fall, 8. (5 points) Find the intervals where the graph of f () = is concave upward and concave downward. Answer. f () = + 7 = (5 7)( ) = at =, = 7 (Critical Numbers) 5 f () = = at =.85 (Candidate for Inflection Point) / Sign of f + Graph of f Concave Downward f (/) = 686/6.88 Concave Upward From the table, we observe f () is concave upward on (.85, ) and concave downward on (,.85). Since the concavity changes around the point (.85,.88), the point (.85,.88) is the inflection point of f (). 5 y y 5 Inflection Point Graph of f () = Graph of f () = + -5 = +. (5 points) Use the Second Derivative Test to find the local etrema of f () = +. Answer. f () = = = at = ± (Critical Numbers) f () =, f ( ) = <, f () = >. By the Second Derivative Test, we conclude f () has the local maimum value f ( ) = and the local minimum value f () =. Page of 5 Points Earned: out of points.

10 Calculus I for Engineering SECOND WORKSHEET SOLUTION Fall, 8 5. (5 points) Find the slant/oblique asymptote of the graph of f () = Answer. We observe f () = = , Thus, f () has the slant/oblique asymptote y =. lim 6 6 [f () ( )] = lim + 7 =. 6. (5 points) Bubba, a farmer, wants to fence in a three sided area net to his barn, using the side of the barn as the fourth side of the rectangle. He also wants to divide this rectangular area into three identical pens (see figure below). Suppose Bubba has 6 feet of available fencing. Find the dimension of each pen which maimize its area. Round dimensions to nearest tenth. Barn Pen Pen Pen Answer. Let and y be the lengths of width and height of one pen. Then each pen has area y and total area of three pens is A = y. Moreover, + y = 6, that is, A = y, + y = 6, < < 6 Putting = 6 y into A = y, we get 5., < y < 6 =. A(y) = y(6 y) = y( y), A (y) = 8(7 y) = at y = 7 (,) (Critical No.) Since A (y) = 8 < for all y, so A (7) = 8 <. By the Second Derivative Test, A(y) has 6 (7) the local maimum value A(7) = (7)( 7) = 56 with = = and y = 7. Since lim A(y) = and lim A(y) =, hence A(7) = 56 is in fact the absolute maimum y y 6/ value. A pen is a small area with a fence round it in which farm animals are kept for a short time. Page of 5 Points Earned: out of points.

11 Calculus I for Engineering SECOND WORKSHEET SOLUTION Fall, 8 7. (5 points) Find the point (a, b) on the curve of y = which is closest to the point (,). Answer. The distance between (,) and the point (a, b) on the curve of y = f () = is d = (a ) + b, d = (a ) + b. Since (a, b) is on the curve, we have b = f (a) = a, i.e., b = a. Putting this into the equation on the distance, we have d = (a ) + a, which we want to minimize with the condition a and b 9. (Why?) Let g(a) = (a ) + a and find the absolute minimum value. g (a) = (a )(a + a + ) = at a = [,] (Critical Number) Since g (a) = (6a + ) > for all a, so g () = >. Hence, by the Second Derivative Test, g() = 5 is the local minimum value with a = and b =. Since the values of g(a) at the endpoints of the interval of a are g() = 9 and g() = 8, hence, g() = 5 is in fact the absolute minimum value, i.e., the point (,) gives the shortest distance d = 5 from the curve to the given point (,). 8. (5 points) Find the volume of the largest bo with no top that can be made from a piece of cardboard 5 inches square by cutting equal squares from the corners and turning up the sides. 5 5 Answer. The bo has the height, while the length and width are 5 so the volume V of the bo is V () = (5 ) = ( 7), < 5, < < 5 = 7. To find the maimum value of V, we find the critical numbers of V (): V () = ( 9)( 7) = at = 9 (,7) (Critical Number) Since V () = ( 8) and V (9) = 6 <, by the Second Derivative Test, V () has the local maimum value V (9) = 66. Because V () = = V (7), the local maimum value is in fact the absolute maimum value, i.e., the volume of the largest bo is V (9) = 66 with the length and width 6 and height 9. Page 5 of 5 Points Earned: out of points.

12 United Arab Emirates University College of Sciences Department of Mathematical Sciences WORKSHEET SOLUTION Chapter Integration Calculus I for Engineering MATH SECTION CRN : :5 on Sunday & Tuesday Due Date: Sunday, November, 8 ID No: Solution Name: Solution Score: 85/85 Instructions:. Read the questions carefully before you start working.. Show all your work to get the full credit.. Make your work to be neat and organized.. Do NOT use the calculator.

13 Calculus I for Engineering THIRD WORKSHEET SOLUTION Fall, 8. (Total 5 points) Find the derivative. (.) (5 points) f (y) = y t sin t dt Answer. By the Fundamental Theorem of Calculus, we get f (y) = d d y y t sin t dt = y sin y u (.) (5 points) y = + u du Answer. By the Fundamental Theorem of Calculus, we get y = d y d = d d 5 (.) (5 points) y = cos(u ) du cos u ( ) du = + u + ( ) ( ) = Answer. By the Fundamental Theorem of Calculus, we get y = d y d = d 5 cos(u ) du d cos ( ) + ( ) = cos [ (5) ] (5) cos [ (cos ) ] (cos ) = 5cos(5 ) + sin cos ( cos ). (5 points) Find a function f and a number a such that for all >, f (t) 6 + a t dt =. Answer. We differentiate both sides of the equation with respect to and simplify the equation: f () =, i.e., f () = /, i.e., f () = / d = 5 5/ + C where C is the constant of integration. Putting = a > into the given equation, we deduce a f (t) 6 + a t dt = a, i.e., 6 = a. Thus, we get a = 9. Page of 5 Points Earned: out of points.

14 Calculus I for Engineering THIRD WORKSHEET SOLUTION Fall, 8. (5 points) Find the interval on which the curve y = f () = dt is concave up- + t + t ward. Answer. We find the second derivative of y = f () = We observe + t + t dt: y = f () = + +, y = f + () = ( + + ). f () > + < <. Therefore, the curve is concave upward on the interval (, ).. (5 points) If f () = and f () is continuous and f ()? f () d = 7, what is the value of Answer. By the Fundamental Theorem of Calculus, we deduce 7 = f () d = f () f () = f (), where f () = is a given condition. Hence, we conclude f () = (5 points) Evaluate the integral + d by making the substitution u = +. Answer. The substitution u = + implies du d =, d = du. The given integral becomes + d = u du = 9 u/ + C = ( + ) / + C, 9 where C is the constant of integration. Page of 5 Points Earned: out of 5 points.

15 Calculus I for Engineering THIRD WORKSHEET SOLUTION Fall, 8 6. (Total 5 points) Evaluate the indefinite integral. (6.) (5 points) 5 d Answer. The substitution u = 5 implies d = du and thus the integral becomes 5 d = u du = ln u + C = ln 5 + C (6.) (5 points) (ln ) d Answer. The substitution u = ln implies d = du and thus the integral becomes (ln ) d = u du = u + C = (ln ) + C (6.) (5 points) cos t t dt Answer. The substitution u = t implies dt t = du and thus the integral becomes cos t dt = t cos u du = sin u + C = sin t + C (6.) (5 points) e + e d Answer. The substitution u = + e implies e d = du and thus the integral becomes (6.5) (5 points) e u + e d = du = u/ + C = ( + e ) / + C + + d Answer. We separate the integral: + + d = + d + + d. The substitution u = + implies d = du and thus the latter integral becomes + d = u du = ln u + C = ln( + ) + C. Page of 5 Points Earned: out of 5 points.

16 Calculus I for Engineering THIRD WORKSHEET SOLUTION Fall, 8 Therefore, we conclude + + d = + d + 7. (Total 5 points) Evaluate the definite integral. (7.) (5 points) e / d + d = tan + ln( + ) + C Answer. The substitution u = implies d = du. The endpoints = and = of the integral correspond to u = and u =, respectively. Thus the definite integral is converted and computed: = e / = d = u=/ u= e u du = [ e u] u=/ u= = e e /. (7.) (5 points) e e d ln Answer. The substitution u = ln implies d = du. The endpoints = e and = e of the integral interval correspond to u = ln e = and u = ln ( e ) = ln e =, respectively. Thus the definite integral is converted and computed: =e =e d u= ln = du = [ u ] u= u= u u= =. (7.) (5 points) a a d, where a is a constant. Answer. The substitution u = a implies d = du. The endpoints = and = a of the integral interval correspond to u = a and u =, respectively. Thus the definite integral is converted and computed: =a = 8. (5 points) If f is continuous and a d = u= u du = [u /] u= u=a = a u=a. f () d =, then find f () d. Answer. The substitution u = implies du = d. The endpoints = and = of the integral interval correspond to u = and u =, respectively. Thus the definite integral is converted and computed: = = f () d = u= u= f (u) du = = 5. Page of 5 Points Earned: out of points.

17 Calculus I for Engineering THIRD WORKSHEET SOLUTION Fall, 8 9. (5 points) If a and b are positive numbers, show that a ( ) b d = ( ) a b d. Proof. We convert the integral in the left hand side of the equation and deduce the integral of the right hand side. The substitution u = implies d = du and = u. The endpoints = and = of the integral interval correspond to u = and u =, respectively. Thus the integral in the left hand side of the equation is converted: = = a ( ) b d = u= u= ( u) a u b du = u= u= ( u) a u b du. We recall that an indefinite integral represents a function, but a definite integral represents a finite number (at least in this course) and so = = This implies d = y= = y d y = y= u= u= k= k= k dk = ( u) a u b du = Math= Math= = = Math d(math) = ( ) a b d, Cal= Cal= Cal d(cal). which is the integral in the right hand side of the given equation. Therefore, we conclude = = a ( ) b d = u= u= ( u) a u b du = = = ( ) a b d. Hence, it is proved. b Remark to Problem Above: Why do we need the positiveness of the constants a and b? If a = and b =, then the integrand a ( ) b becomes, which has the ( ) vertical asymptotes = and =. That is, the definite integral dose not eist: a ( ) b d =. b Application: Evaluate the integral Answer. By the equation proved above, we have ( ) 8 d = ( ) 8 d = = [ 9 9 ( ) 8 d. ] ( 8 9) d = 9 = Page 5 of 5 Points Earned: out of 5 points.

18 United Arab Emirates University College of Sciences Department of Mathematical Sciences WORKSHEET SOLUTION Chapter 6 Techniques of Integration Calculus I for Engineering MATH SECTION CRN : :5 on Sunday & Tuesday Due Date: Sunday, November 6, 8 ID No: Solution Name: Solution Score: 55/55 Instructions:. Read the questions carefully before you start working.. Show all your work to get the full credit.. Make your work to be neat and organized.. Do NOT use the calculator.

19 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8. (Total points) Evaluate the following integrals. (.) (5 points) + d Method. Let u =. Then The integral becomes + d = du d = =, i.e., d = u du. u ( u + u du = ) du + u (by Division Algorithm) = u ln + u + C = ln + + C. Method Better Choice. Let u = +. Then The integral becomes du d = =, i.e., d = (u ) du. (u ) ( (u ) + d = du = ) du u u = u ln u + C = ( + ) ln + + C = ln + + C. (.) (5 points) ln d Method. Let f () = and g() = ln. Then we get f () = / and g () =. By the Integration by Parts formula, we have ln d = / ln ( ) d = / ln / + C = [ / ln ] + C. Method Advanced. Let u = /. Then we get du d = = u/, i.e., d = du. u/ Page of Points Earned: out of points.

20 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8 The integral becomes (.) (5 points) ln d = u / ln u / du u/ ( ) ( ) = ln u du = u [ln u ] + C (by Eample in Tetbook) ( ) [ = / ln / ] + C = [ / ln ] + C. e 8 6 d Method Trigonometric Substitution. Let e = secθ, where θ [,π/) (, because e is always positive). Then e 8 6 = (secθ) 6 = 6sec θ 6 = 6(sec θ ) = 6tan θ = (tanθ), e 8 6 = = (tanθ) tanθ, d d e = d d (secθ), secθ d = secθ tanθ dθ, i.e., d = tanθ dθ. e = secθ tanθ dθ d, i.e., e d = secθ tanθ dθ, The integral becomes e 8 6 d = tanθ tanθ dθ = 6 dθ = 6 θ + C We epress the result in terms of. Since e = secθ, so the triangle with the angle θ implies secθ = e, cosθ =, sinθ = e e 8 6 e 8 6 e, tanθ =. It gives us ( ) θ = cos e ( ) e and θ = sin 8 6 e ( ) e and θ = tan 8 6. Using the last one (arctangent: one may use arccosine or arcsine result), finally, we conclude e 8 6 d = 6 θ + C = ( ) e tan + C. Method Another Substitution. Page of Points Earned: out of 5 points.

21 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8 Let u = e 8 6. Then u = e 8 6 and we get u du d = 8e8 = 8(u u + 6), i.e., d = 8(u + 6) du = u (u + 6) du. The integral becomes e 8 6 d = u u (u + 6) du = u + 6 du = 6 [ (u/) + ] du = 6 w + dw (Substitution w = u/) = 6 tan w + C = ( u ) 6 tan + C = ( ) e tan + C. Method One More Substitution. We observe ( e e 8 8 ) ( e 8 ) ( (e 6 = 6 6 = 6 ) ) [ ( e = 6 = ) ], e 8 6 = [ ( e ) ] = ( ). e ( ) e ( e Let u =. Then u ) = and we get u du ( e d = 8 ) = 8(u + ), i.e., d = u 8(u + ) du = u (u + ) du. The integral becomes e 8 6 d = u u (u + ) du = 6 u + du (.) (5 points) ( + ) d = 6 tan u + C = 6 tan ( ) e + C. Answer. First, we find the partial fraction decomposition of the integrand: ( + ) = A + B + C +, Page of Points Earned: out of 5 points.

22 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8 where A, B and C are constants to be determined. equation by ( + ), we get Multiplying both sides of the = A( + ) + (B + C). Putting =, = and =, we deduce A =, B = and C =, respectively, i.e., ( + ) = + +, + ( + ) d = d + d = ln + d + d = ln ln + tan ( ) + C. To deduce the last equality above, the substitution u = / was used. (.5) (5 points) 5 d Answer. Let = 5secθ, where θ [,π/) (π/,π]. Then 5 = 5sec θ 5 = 5 ( sec θ ) = 5tan θ = (5tanθ) 5 = (5tanθ) = 5tanθ d = 5secθ tanθ, i.e., d = 5secθ tanθ dθ. dθ So the integral becomes 5 d = = 5 (5secθ) 5tanθ 5secθ tanθ dθ = 5 cosθ dθ = sinθ + C. 5 secθ dθ We epress the result in terms of. Since = 5secθ, so the triangle with the angle θ implies secθ = 5, cosθ = 5, and sinθ = 5. Therefore, we conclude (.6) (5 points) 5 d = 5 sinθ + C = 5 + C. 5 d (Hint: Verte Form of Quadratic Function) + + Answer. First, we start with the verte form of the quadratic function f () = ++ Page of Points Earned: out of points.

23 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8. A simple computation shows f () = + = ( + ) = at = and f ( ) = 9. That is, the verte of the graph of f () is (,9). It implies the verte form of f (): Now we modify the function: f () = + + = ( + ) + 9. [ ( + ) + + = ( + ) ] [( ) = 9 + = + ] = [ ( + ) ] = ( + + ) d = ( d. + ) + Let u = +. Then we get du d =, i.e., du = d. It implies + + d = ( d = + ) + u + du = u + du. To solve the last integral, we use the trigonometric substitution u = tanθ. u + = tan θ + = sec θ, du The integral becomes u + = = sec θ secθ dθ = d dθ tanθ = sec θ, i.e., du = sec θ dθ. + + d = = u + du = secθ sec θ dθ secθ dθ = ln secθ + tanθ + C, by the eample in the tetbook. Now we epress the result in terms of. Since u = + and u = tanθ, so by the Page 5 of Points Earned: out of points.

24 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8 triangle with the angle θ, we have tanθ = +, sinθ = + ( + ) + = + + +, cosθ = = ( + ) + + +, secθ = + +. Putting them into the result above, finally we conclude + + d = ln secθ + tanθ + C = ln C = ln + C = ln ln + C = ln C. Page 6 of Points Earned: out of points.

25 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8. (5 points) Let Evaluate the integral 7 f () = f () d. + + for 6, for > 6. Answer. Since the function is defined piecewisely, we separate the integral as follows: f () d = f () d + f () d = 6 + d + 6 d = I + II, + where I and II represent the first and second integral, respectively and we evaluate each integral. Part A. Integral I = 6 + d: The substitution u = + implies du d =, i.e., du = d. The endpoint of the interval, = and = 6, correspond to u = and u = 7, respectively. So the integral becomes 6 I = + d = 7 u du = 7 Part B. Integral II = 6 + d: A simple observation tell us (or by Division Algorithm) 7 6 [ ] 7 ln u = ln7. + = + + = = + 7 ( + d = ) [ ] 7 d = tan + = 6 tan 7 + tan 6. 6 Therefore, the sum of I and II is f () d = + d d = I + II = ln7 + tan 7 + tan 6. Page 7 of Points Earned: out of 5 points.

26 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8. (5 points) Suppose f () and f () are continuous. Show that f ()f ()e f () d = e f () ( + f ()) + C, where C is any constant. Method Substitution and Integration by Parts Formula. Let u = f (). Then we have du d = f (), i.e., du = f () d. So the integral on the left hand side of the given equation becomes f ()f ()e f () d = ue u du. The Integration by Parts formula with h (u) = e u and k(u) = u implies h(u) = e u and k (u) = and thus f ()f ()e f () d = ue u du = h (u)k(u) du = h(u)k(u) h(u)k (u) du = ue u + e u du = ue u e u + C = e u (u + ) + C = e f () (f () + ) + C. Method Differentiation. We recall that if G () = g(), then G() = side of the given equation: d d g() d. So we differentiate the right hand [ ] [ e f () ( + f ()) + C = e f ()] ( + f ()) e f () [ + f ()] (Product Rule) = e f () f ()( + f ()) e f () f () (Chain Rule) = e f () f () + e f () f ()f () e f () f () = e f () f ()f (). Hence, it implies e f () ( + f ()) + C = e f () f ()f () d. Page 8 of Points Earned: out of 5 points.

27 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8. (5 points) In an AC circuit, the current has the form I(t) = A sin (ωt) for constants A and ω. The power is defined [ as I (t). Find the average value of the power I (t) by integrating it over the interval, π ]. (Hint: Average value of a function over an interval was ω discussed in Section..) Answer. By the definition on the function average, the average value of I (t) over [,π/ω] is obtained by π/ω (I I (t) dt (t)) ave = = ω π/ω I (t) dt π/ω π = ω π/ω A sin (ωt) dt = ωa π/ω ( sin (ωt) ) dt π π = ωa π/ω ( ) cos(ωt) dt (Half Angle Formula on sin (ωt)) π = ωa π/ω cos(ωt) + cos (ωt) dt π = ωa π/ω ( cos(ωt) + cos (ωt) ) dt 8π { [ = ωa t sin(ωt) ] π/ω } π/ω + cos (ωt) dt 8π ω { = ωa π π/ω } 8π ω + + cos(ωt) dt (Half Angle Formula on cos (ωt)) { = ωa π 8π ω + π/ω } ( + cos(ωt)) dt { = ωa π 8π ω + [ t + sin(ωt) ] } π/ω { = ωa π ω 8π ω + } π = A ω 8. Therefore, the average value of the power I (t) is A 8. Page 9 of Points Earned: out of 5 points.

28 Calculus I for Engineering FOURTH WORKSHEET SOLUTION Fall, 8 5. (5 points) Among the following indefinite integrals listed below, evaluate the ones that can be worked out using the techniques of integration learned in class. e A. e d B. e d C. e d D. e d E. d Answer. A. Cannot be solved. B. Integration by Substitution technique works good with u =. e d = e + C. C. Cannot be solved. D. The Integration by Parts formula can be used with f () = e and g() =. e d = e ( ) + C. E. Cannot be solved. b Aside: Can you see the pattern above? What does your intuition tell you for the integrals? (i) positive even integer e d and (ii) positive odd integer e d We can say that the integral of (i) cannot be solved. On the other hand, the integral of (ii) can be solved and there should be a formula (involved with the Gamma function). What is the formula and how can we find it? If you have these questions in your mind, you must major in mathematics. You are born to be a mathematician. 6. (5 points) Suppose that f () and g() are continuous functions satisfying g() = f () d, g() =, g(8) = 6. Find f () d. Answer. Let u =. Then du d =, i.e., d = du and = and = correspond to u = and u = 8, respectively. So the integral becomes = f () d = f () d = u=8 f (u) du = 6 + [g(8) g()] = = u= = 5. Page of Points Earned: out of points.

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