Math 261 Final Exam - Practice Problem Solutions. 1. A function f is graphed below.

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1 Math Final Eam - Practice Problem Solutions. A function f is graphed below. f() (a) Find f(), f( ), f(), and f() f() = ;f( ).;f() is undefined; f() = (b) Find the domain and range of f Domain: (,) (, ) Range: [, ) (c) Find the intervals where f () is positive f () > on (,) (, ) (d) Find the intervals where f () is negative. f () < on (,) (,) (, ) (e) Find lim f() lim f(). (f) find lim f() and lim f() + lim f() = ; lim f() = + (g) find lim f() and lim +f() f() = lim +f().9 lim (h) find lim f() and lim f() lim f() = and lim f() = (i) find the points where f() is discontinuous, and classify each point of discontinuity. f has a removable discontinuity at =, an infinite discontinuity at =, and a jump discontinuity at =.. Evaluate the following limits: (a) lim = lim (+)( ) (+)( ) = (+) (+) = 8 (b) lim = lim (+)( ) (+)( ) = (+) (+) = 7 = (c) lim is undefined since is only defined for. (d) lim cos = π (e) lim cos is undefined (cos continues to oscillate from to and back) (f) lim = (g) lim =

2 . Given the function + if f() = if < if (a) Graph f(). f() (b) Find lim lim f(), lim +f(), and lim f() f() =, lim +f() =, and lim f() = (c) Is f() continuous at =? Justify your answer. Yes. In an interval containing =, the function f is defined by which is a polynomial and hence is continuous.. Give the formal ǫ - δ definition of the limit of a function as presented in class. Then draw a diagram illustrating the definition. Finally, write the definition informally in your own words. Let f be defined on an open interval containing a, ecept possibly at a itself, and let L be a real number. The statement f() = L means that for every ǫ > there is a δ > such that if < a < δ, then f() L < ǫ. lim a f() L + ε L ε y = L a +δ a a δ Intuitively, what the formal definition of a limit says is that lim a f() = L means that if we set an error tolerance of ǫ on the y-ais, then no matter how small we set our error tolerance, it is possible to choose an error tolerance δ on the -ais so that all points within δ of a on the -ais get mapped by the function to points that are within ǫ of our limit value L.. Given that f() =, lim f() =, and ǫ =., find the largest δ such that if < < δ, then f() < ǫ. We need f() < ǫ. That is <. or <. Therefore,. < <., or.99 < <., so.99 < < Hence < < or, rounding,.9989 < <.8.. Thus.8 < <.8. So we can take δ <.8

3 . Use the formal definition of a limit to prove that lim =. Suppose f() L < ǫ. Then < ǫ or < ǫ. That is, ( ) < ǫ, or < ǫ, so < ǫ. Let δ ǫ. Then if < < δ ǫ, < ǫ. Therefore = = < ǫ. That is, f() < ǫ. Hence lim f() = 7. Let f() =. (a) Find the values of at which f is discontinuous. Notice that f() = = ( ) ( )(+). Since f is a rational function, it is continuous everywhere ecept where the denominator is zero. Therefore, f is discontinuous at = and at =. (b) Find all vertical and horizontal asymptotes of f. Since the discontinuity at = is removable, f only has one vertical asymptote at =. Since lim f() =, f has a horizontal asymptote at y =. 8. Find the values at which f() = + + is continuous. Since f consists of a sum of square roots of polynomial functions, f is continuous wherever it is defined. Thus we need, or and we need + >, or >. Thus f is continuous on the interval (, ]. 9. (a) Use the Intermediate Value Theorem to show f() = + has a root between and. First notice that f is a polynomial, and hence is continuous everywhere, so the IVT applies in this situation. Net, f() =, and f() =, so by the IVT, f attains every value between - and at least once for some -value in the interval (,). In particular, there must be some c in the interval (,) with f(c) =. (b) Use Newton s method to approimate this root to decimal places. Recall that Newton s method uses the derivative to recursively approimate a root of a function. Given an initial guess, we compute approimations using the formula: n+ = n f(n) f ( n) Here, f () = +, and we will take =.. Then = ; = , =.87979, =.87988, =.87988, so our approimation of the root to four decimal places is : = Find the derivative y = dy d (a) y = π +π π + π y = π π +π π (b) y = cos()+sin() y = sin()+cos() (c) y = +cos( ) y = sin( ) (d) y = tan y = tan+ sec () for each of the following: (e) y = sec ( ) y = sec ( ) sec( )tan( ) = 9 sec ( )tan( ) (f) y = + (g) y = y = ()( +) ( )() ( +) = ( +). cos +sin( ) y = (cos sin)(+sin( )) ( cos)( cos( )) (+sin( )) (h) y = sin (tan( )) y = sin(tan( ))cos(tan( ))sec ( )

4 (i) y +y = Differentiating implicitly, y y +yy =, so yy y = y, or y (y ) = y. Hence y = y y (j) y y y = Differentiating implicitly, y + y y y yy =, so y y = y + y +yy = y ( ++y). Hence y = y y ++y. Use the formal limit definition of the derivative to find the derivative of the following: (a) f() = + f (+h) (+h)+ ( +) +h+h h + () = lim = lim h h h h h+h h = lim = lim +h = h h h (b) f() = f +h () = lim h h h+ = lim h (+h )( ) h h = lim h (+h )( ) = lim h h = lim h ( ) (+h ) (+h )( ) h (+h )( ) = (c) f() = + +h+ + +h++ + f () = lim h h +h++ + h = lim h +h+ (+) h( +h++ +) = lim h = lim = h +h ( ) h( +h++ +). Use the quotient rule to derive the formula for the derivative of sec(). Recall that sec = cos, so by the quotient rule: d d sec = d d cos = ( sin) cos = sin cos = cos sin cos = sectan.. Given that h() = f(g()) and the following table of values: f() g() f () g () Find the following: (a) (f +g)() = f()+g() = 7+( ) = (b) (f +g) () = f ()+g () = + = (c) (fg)() = f() g() = = (d) (fg) () = f ()g()+f()g () = +()() = ( ) (e) f g () = f() g() = 7 (f) h () = f (g())g () = f ()g () = ( )() =. If f() =, find the intervals where f() is continuous, and find the intervals where f() is differentiable. First notice that f is continuous wherever it is defined, so we need, or, so f is continuous on (, ]. Net, f () = ( ) ( ) = ( ) = Therefore, f is differentiable when >, or when <. That is, on the interval (, ).

5 . The total net value of two companies: Company A and Company B, in millions of dollars as a function of time (in years since the year ) is given by f(t) and g(t) respectively. 8 g(t) f(t) 7 t (a) How fast was company A s value growing in? (include units for your answer) Notice that the growth of company A in is given by the value of the derivative of f when t =, but since f is linear, its derivative is constant, so we have f () = 7 7. in millions of dollars per year. (b) What was the average rate of change for company B from through 7? This is given by the slope of the secant line between (,g()) and (7,g(7)). Here m av = 7 = 7.87 in millions of dollars per year. (c) Which company s value was growing faster in? This is a bit open to interpretation since there is a cusp point on g(t) when t =. Looking backwards, the value of company B is increasing more steeply than that of company A, but the rate of change of company B switches from positive to negative at t =. Since we usually view time as moving forward, we will say that at the beginning of, Company B s value is growing faster. (d) Which company s value was growing faster in? The situation here is much clearer. Looking at the tangent lines to each of the functions when t =, we see that the tangent line to f(t) has larger slope, so Company A s value is growing faster at the beginning of. (e) Which company would you rather own stock in, and why? This question is a bit qualitative, since we cannot know what will happen in the future for thee companies, but it appears that Company A is steadily increasing in value, while Company B has topped out and is starting to decline. I would rather own stock in Company A.. Find the equation for the tangent line to f() = when =. Then, use this tangent line to approimate. First notice that f () =, so the slope of the tangent line is ( ) ( ) = (8) = =. Net, notice that f() = = 8 =, so by point/slope, the tangent line is given by the equation y = ( ), or y = +. Now, f() = =. Then, using our tangent line to approimate: ()+ = + =.7. (In fact, calculating directly,.9) 7. Find the equation of the tangent line to the graph of f() = sin() when = π f () = cos(), so m = cos ( ) π = = Also, f( π ) = sin( π ) = Therefore, the tangent line equation is given by y = π, or y = + π 8. Find the tangent line to the graph of the relation y = y at the point (,) Differentiating implicitly, y +y 8 = yy, so y 8 = yy y = y (y ). Thus y = y 8 y, which when = and y = gives:y = () 8 () = uh oh! The tangent line is undefined! There is no tangent line at this point.

6 9. Prove the Quotient Rule by applying the the Chain Rule to the general function h() = f()[g()]. If h() = f()[g()], then h () = f ()[g()] +f() ( )[g()] g () = f () g() f()g () g() = f ()g() g() f()g () g() = f ()g() f()g () g(). A person flying a kite holds the string feet above ground level, and the string is played out at a rate of feet per second as the kite moves horizontally at an altitude of feet. Assuming that there is no sag in the string, find the rate at which the kite is moving when a total of feet of string has been let out. Let L be the length of the string. Let be the horizontal distance of the kite from the place where the person is standing. Then, since the net height of the kite is feet ( - feet), using the Pythagorean Theorem, we have L = +. [Note: since the motion is horizontal, y is constant] Differentiating this implicitly, LL =, or solving for, = LL = LL. Notice that L =, L =, and = () = 7 at the time in question. Therefore, = ()() 7 = ft sec.. Draw the graph of a function f() that is continuous when =, but is not differentiable when =. f() Find f () () if f() = cos() f () = sin(); f () = cos(); f () () = 8sin(); f () () = cos(); f () () = sin(); f () () = cos() = cos(). Find f () () if f() = +7 f () () =. Each time you differentiate, the power drops by, so the th derivative is a constant. Hence, the st derivative is zero.. The graph of f is given. Answer the following questions. f ()

7 (a) Find the intervals where f is increasing. f is increasing whenever f is positive. That is, on the intervals: (,) (,7). (b) Find the intervals where f is decreasing. f is increasing whenever f is negative. That is, on the intervals: (,) (7, ). (c) Find the location of all local maimums. There are local maima whenever f goes from positive to negative, so at = and = 7 (d) Find the location of all local minimums. There are local minima whenever f goes from negative to positive, so at = (e) Find the intervals where f is concave up. f is concave up whenever f is increasing. That is, on the interval: (,) (f) Find the intervals where f is concave down. f is concave down whenever f is decreasing. That is, on the intervals: (,) (, ). (g) Find any inflection points. Inflection points occur when the f changes concavity, so at: = and at = (h) Sketch a possible graph for f(). f() (i) Find the location of the absolute maimum on [,7], if one eists. The absolute ma is when = 7. We see this by looking at areas under f () and convincing ourselves that f(7) must be greater than f() and f(). (j) Find the location of the absolute minimum on [,7], if one eists. The absolute min is when = (this requires us to notice that the amount added to f() as goes from to is less than the amount subtracted from f() as goes from to, so f() > f()). (k) Sketch a possible graph of f f "()

8 . Find the absolute maimum and the absolute minimum of each of the following functions on the given interval. (a) f() = + on the interval [,7] First notice that f is continuous so the EVT applies. Net, f () = = when = Now f( ) = ; f() =, and f(7) = 9, so the absolute ma is: and the absolute min is: - (b) f() = +7 on the interval [,] First notice that f is continuous so the EVT applies. Net, f () = = when =, but this is not in our interval of interest. Now f( ) = and f() =, so the absolute ma is: and the absolute min is: - (c) f() = ++ on the interval [,] First notice that f is continuous so the EVT applies. Net, f () = + = when = or =. Now f() = ; f() = 9, f() = and f() = 9, so the absolute ma is: 9 and the absolute min is:. (d) f() = + on the interval [,] First notice that f is continuous on [,] (the only value we need to worry about is = ) so the EVT applies. Net, f () = ( ) (+) ( ) = ( ) which is never zero. Now f() =, and f() = 7, so the absolute ma is: and the absolute min is: 7. (e) f() = on the interval [,] Notice that f is discontinuous at =, so the EVT does not apply. In fact, the function has a vertical asymptote that attains values approaching on one side and on the other side, the this function has no absolute ma and no absolute min on this interval.. Sketch a graph which satisfies the following: domain is { } y-intercept is (, ) -intercepts are (,), (,), and (7,) the vertical asymptote is at = lim f() = lim f() = increasing on (, ) (, ) decreasing on (, ) (oops - typo on original handout!) local min at (, ) no local maima concave up on (, ) (,) (,7) concave down on (,) (7, ) inflection points at (, ),(, ),(7,) = f() y = y =

9 7. A particle is moving along a straight line during a sity-second time interval. Every ten seconds the position, velocity, and acceleration of the particle are measured, and recorded in the table below. For each of the following questions, answer the question and justify your answer. t (seconds) s(t) (feet) 8 v(t) (ft/s) a(t) (ft/s ) (a) Is there a time when s(t) = feet? Yes. We may apply the IVT to the function s(t) over the interval [,]. Since s() = < < s() = 8 we may conclude that there is a c in the interval (,) with s(c) = feet. (b) Is there a time when v(t) = ft/sec? Yes. We may apply the IVT to the function v(t) over the interval [,]. Since v() = < < v() = we may conclude that there is a c in the interval (,) with v(c) = ft/sec. (c) Is there a time when v(t) =. ft/sec? Yes. We may apply the IVT to the function v(t) over the interval [,]. Since v() = <. < v() = we may conclude that there is a c in the interval (,) with v(c) =. ft/sec. (d) Is there a time when v(t) = ft/sec? Since v(t) for all values in the table, the IVT will not help us here. However, we may apply the MVT to the function s(t) over the interval [,]. Since s() s() = 8 = we may conclude that there is a c in the interval (,) with s (c) = v(c) = ft/sec. (e) What is the largest velocity that you can justify from the data in the table? The largest value we can conclude using the IVT is 8 ft/sec. However, we may apply the MVT to the function s(t) over the interval [,]. Since s() s() = 8 =. we may conclude that there is a c in the interval (,) with s (c) = v(c) =. ft/sec. (f) Is there a time when a(t) = ft/sec? Yes. We may apply the IVT to the function a(t) over the interval [,]. Since a() = < < a() = we may conclude that there is a c in the interval (,) with v(c) =. ft/sec. (g) Is there a time when a(t) =. ft/sec? Since a(t) for all values in the table, the IVT will not help us here. However, we may apply the MVT to the function v(t) over the interval [,]. Since v() v() = =. we may conclude that there is a c in the interval (,) with v (c) = a(c) =. ft/sec. 8. Find a number c in the given interval that satisfies the Mean Value Theorem for the function and interval given, or eplain why the Mean Value Theorem does not apply. (a) f() = + on [,] f() f( ) First, notice that f is continuous on [,] so the MVT applies. Also, = ( ) =. Net, ( ) f () = +, so we need to find so that + =. That is, =, or =. Thus = is the value that satisfies the Mean Value Theorem on the interval [, ]. (b) f() = on [,] Here, notice that f() has a vertical asymptote when =, thus the MVT does not apply. 9. A hotel that charges $8 per day for a room gives special rates to organizations that reserve between and rooms. If more than rooms are reserved, the charge per room is decreased by $ times the number of rooms over. Under these terms, what number of rooms gives the maimum income? Let be the number of rooms reserved. We know, but since there is no discount for below rooms, we know that. The price per room for between and rooms is given by: p() = 8 ( ) =. Therefore, the revenue from rooms is given by R() = ( ) =. Now R () = = when =, or when =. Thus, applying the EVT, since R() =, R() = and R() =, so the maimum revenue occurs when rooms are reserved.

10 . A page of a book is to have an area of 9 square inches with inch margins on the bottom and sides of the page and a /-inch margin at the top. Find the dimensions of the page that would allow the largest printed area. Let be the width of the page and y the length (in inches). Then y = 9, or y = 9. Accounting for the margins, we canseethattheareaofpartofthepageusedforprintingisgivenby: A = ( )(y ) = ( )(9 ) = 9 8. Therefore, A () = 8 8 = when =, or = =, so =. Notice that A () = < for >, so this is a maimum. Therefore the page should have dimensions =.9 inches and y = 9 8. inches.. Sketch the following functions using information about the domain, the intercepts, the intervals where the function is increasing or decreasing, any local maimum or minimum, the intervals where it is concave up or concave down, any inflection points, and any asymptotes. (a) f() = 8 Note that f () = 8 = ( )(+) and f () = = ( ). Doing a full workup using the first and second derivative tests yields the following results: Domain (, ) y-intercept (, ) ( + -intercepts (, ), ) (,, ), Decreasing: (, ) Increasing: (, ) (, ) Local Maimum at (,); Local Minimum at (, ) Concave up: (, ) Concave down: (,) Inflection Point at (, ) No Asymptotes (b) f() = Note that f () = ( ) and f () = ( ). Doing a full workup using the first and second derivative tests yields the following results: Domain ( ) (,, ) y-intercept (, ) -intercept (, ) Decreasing: ( ) (,, ) Increasing: None No Local Etrema Concave up: (, ) Concave down: ( ), No Inflection Points Asymptotes: Vertical at =. Horizontal at y = ( lim = ).

11 (c) f() = sin+cos Note that f () = cos sin and f () = sin cos. Doing a full workup using the first and second derivative tests yields the following results: Domain (, ) y-intercept (, ) -intercepts ( π +kπ,) for any integer k. Decreasing: ( π +kπ, π +kπ) for any integer k. Increasing: ( π +kπ, 9π +kπ) for any integer k. Local Maima: ( π +kπ, ) for any integer k. Local Minima: ( π +kπ, ) for any integer k. Concave up: ( π +kπ, 7π +kπ) for any integer k. Concave down: ( 7π +kπ, π +kπ) for any integer k. Inflection Points: ( π +kπ,) for any integer k. Asymptotes: None.. Suppose you are standing on the roof of a shed that is feet tall and you throw a ball vertically upward. If you release the ball feet above the top of the shed at an initial speed of feet per second, how high will the ball travel and when will the ball hit the ground? (Assume gravity is ft/sec ) From the description above, we see that s() =, v() =, and a(t) =. Then v(t) = t+, and s(t) = t +t+. The ball will attain its maimum height when v(t) =, that is when t =, or when t =. Then the maimum height is s( ) = ( ) +( )+ = ++ = 9 feet. The ball will hit the ground when s(t) =, that is, when t +t+ = 8(t t ) = 8(t+)(t ) =, which has solutions t = and t =. Clearly the negative solution does not make sense, so the ball must hit the ground after seconds.

12 . Use Newton s Method to approimate a real root of the function f() = + to decimal places. A clever student will notice that f() =, but that is not really using Newton s method, so we will pretend that we didn t notice this and apply Newton s method. Recall that Newton s method uses the derivative to recursively approimate a root of a function. Given an initial guess, we compute approimations using the formula: n+ = n f(n) f ( n) Here, f () =, notice that f () = when = and when so we need to avoid these values (why?). We will take =.. Then = ; =.99, = , = (we are probably rounding too soon here, but that s all the precision my calculator has), =,..., so our approimation of the root to five decimal places is : =... Use Newton s Method to approimate to decimal places. The first trick here is to come up with a polynomial function that has as a root (well, I suppose it wouldn t have to be a polynomial function, but they are pretty easy to come by). If we start with =, then, cubing each side, we have =, so we may take f() =. Then f () =. Newton s method uses the derivative to recursively approimate a root of a function. Given an initial guess, we compute approimations using the formula: n+ = n f(n) f ( n) Since we know that < <, then we will take =. Then =.8; =.87, =.87, =.879, =.879, so our approimation of the root to five decimal places is : =.87.. Find each of the following indefinite integrals: (a) +Cd = +C+D (b) +CdC = C + C +D (c) (d) + d d = + = + +C (e) (f) Let u =. Then du = d, so du = d. This gives the integral u du = u +C = 9 u +C = 9 ( ) +C. tan sec d Let u = tan. Then du = sec d. This gives the integral u du = u +C = tan +C. d Let u =. Then du = d or du = d. Solving u = for, we get u+ = u + =. This allows us to rewrite the integral as u + u du =. Solve the following differential equations under the given initial conditions: u + u du = u + u +C. (a) f () = cos+;f() = Antidifferentiating, we have f() = sin+ +C, so f() = = sin+ +C = C, so C =. Thus f() = sin+ + (b) g () = cos() sin();g (π) = ;g() = Antidifferentiating, we have g () = sin() + cos() + C, so g (π) = sin(π) + cos(π) + C = + C. Therefore, C =, and g () = sin()+cos()+ Net, g() = cos()+ sin()++d, so g() = = cos()+ sin()+()+d = +D. Thus D = 7 Hence g() = cos()+ sin()++7

13 7. Epress the following in summation notation: (a) = k k= 7 (b) = k + k + 8. Evaluate the following sums: k= (a) (b) k(k ) k= = (8 )+(8 )+( )+( ) = ++8+ = 9. (or you can do these using summation formulas, you decide which is easier) k= k k Here, it is probably better to use summation formulas, although you could certainly just calculate and then add up the terms. n ( ) (n)(n+) n ( ) n(n+)(n+) Recall that k = and k =. k= k= [ (()() ) ( ) ] [ (() ) ( ) ] ()() ()(7) Then k k = k k k k = k= k= k= = [ () (8) ] [ 8 ] =,9 9. Epress the following sums in terms of n: (a) (b) n k k + k= = ( n(n+) n k( k ) = ) ( ) n(n+) +n = n +n +n n n +n = n + n n + 7n n n k k = k k ( ) ( 8) ( ) ( ) ( n(n+) n(n+) n = + = + n ) ( ) n + n + n = n n + n + n. Consider f() = in the interval [,] (a) Find a summation formula that gives an estimate the definite integral of f on [,] using n equal width rectangles and using left hand endpoints to give the height of each rectangle. You do not have to evaluate the sum or find the eact area. First notice that = n = n. n n Then A = f(+(k ) ) = = n k= [ k= ( n(k ) + + n (k ) 9 k= ) [(+(k ) n ) ] n ] n n = k= (b) Find the norm of the partition P : < <. < <. < The norm is the width of the largest subinterval, or. [ 8n(k ) 8+ + n k n k +n ] 9 n (c) Find the approimation of the definite integral of f on [,] using the Riemann sum for the partition P given in part (b). Since we are using rectangles with left hand endpoints, the sum is given by: A ()(f())+(.)(f())+(.)(f(.))+(.)(f())+(.)(f(.)) = ()+(7.)+(.7)+(.)+(8.7) =.

14 . Assume f is continuous on [, ], f() d =, (a) (b) (c) (d) f() d = f() d = = f() d = = f() d = = (e) Find the average value of f on [,] The average of f on [,] is given by:. Evaluate the following: f() d =, and f()d = ( ) =. f() d =. Find: (a) (b) (c) (d) + + d = + + = (+ +8) (+ ( +) d +) = 7.7 Let u = + Then du = d, or du = d. Also, + = and + = This gives the integral: π π cos ()sin() d u du = u = 9 9 = 9 9. Let u = cos() Then du = sin()d, or du = sin()d. Also, cos(π ) = and cos(π ) =. This gives the integral: π π sin d Notice that π π u du = sin d = since sin is odd.. Compute the following: ( ) (a) d t dt d t (b) ( ) d t dt dt t (a) Using the first part of the fundamental theorem of calculus, = = (b) Again applying the the fundamental theorem of calculus, (c) d ( ) t dt d t (c) Since (d) [ d dt (d) t t dt is a constant, ( ) t dt t ] = [ ( ) ] d t dt dt t ( d d =. ) t dt = t ( ) d t dt = d ( F( ) F() ) d t d ( ) d t t dt = dt t t

15 . ( points) Find the area of the region bounded by the graph y = sin(), the -ais and the lines = π and = π. A = π π sin d = cos π π = [ ( π cos(π) cos = )] ( ) =. Find the area of the region bounded by the graph = y y and the y-ais. square units. Notice that if we set y y =, then y( y ) =, or y(+y)( y) =. Therefore, y =, y =, or y =. Net, notice that If y =, then () = = >. While if y =, then ( ) ( ) = + = <, so we need to set up to different integrals in order to get the area. Then A = = y y () (y y ) dy + +y y y y () dy = y y dy + = () ( ()())+(() ) () = ( +8)+(8 ) = 8 units. Given the following graph of f(): f() y y dy (a) Approimate rectangle. f()d using 8 equal width rectangles with height given by the right hand endpoint of each Notice that n = 8, and = ( ) 8 = 8 8 =. Also, we are using Right Hand endpoints. Therefore, A [f( )+f( )+f()+f()+f()+f()+f()+f()] = [ ] = (b) Use the Trapezoid Rule with n = to approimate f()d. Notice that n =, so b a n = ( ) = 8 8 =. Therefore, A [f( )+f( )+f()+f()+f()] = [+()+()+()+] = 9 7. (a) Use the Trapezoidal Rule with n = to approimate π sin d Notice that n =, so b a n = π () = π 8. Therefore,A π 8 [sin+f sin π +f sin π +f sin π +f sinπ] = π[ ] = ( ) π (b) Find the maimum possible error in your approimation from part (a). Notice that f () = cos and f () = sin, so M =. The maimum error when using the Trapezoidal Rule is given by: Error M(b a) n = (π) () = π 9.87

16 (c) Find the minimum number of rectangles that should be used to guarantee an approimation of within decimal places using the Trapezoidal Rule. Here, we want Error.. Therefore, we need (π) n. or n 9.7, so the minimum number or rectangles is n = 9. (d) Use the Fundamental Theorem of Calculus to find π does the actual error compare to the maimum possible error? Using the FTC, π sin d = cos π = ( ( ) ( )) =. Our estimate was approimately.88 so the error is about.88 =. The estimated maimum error π sin d to (π) (.) n. Hence n. or sin d eactly. How far off was your estimate? How