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MTH 252 – Calculus II – Fall 2006 Final – 12.13.2006 y f 1 x 1 1. The graph of f (x) = −x2 + 4x + 2 is shown. Answer the following: (7 points each) 4 (a) Estimate f (x) dx with 4 subintervals using left endpoints. 0 4 (b) Evaluate f (x) dx using the Fundamental Theorem of Calculus. 0 2. (Refer to your notes on the “Lanchester’s Laws” project.) In Project #2, Admiral Nelson devised a strategy by which a British ﬂeet was able to defeat a combined French-Spanish ﬂeet with superior numbers. Nelson’s strategy was as follows: Skirmish I: send 32 British ships against 23 opposing ships (the British ﬂeet prevails with 22 ships remaining) Skirmish II: send 8 British ships against 23 opposing ships (the opposing ﬂeet prevails with 21 ships remaining) In the project, we saw that after these two skirmishes, Nelson sent the remaining British ﬂeet against the remaining French-Spanish ﬂeet, and the British ﬂeet was victorious with 6 ships remaining. Let’s play out a diﬀerent scenario. After the two skirmishes, with 22 British ships and 21 opposing ships remaining, suppose that the commandants of the opposing ﬂeet decide to use Admiral Nelson’s strategy against him. They propose the following: Skirmish “A”: send 12 French-Spanish ships against 11 British ships Skirmish “B”: send 9 French-Spanish ships against 11 British ships For each of these new skirmishes, use the general solution y 2 − x2 = C to determine the victor. You won’t need to change the diﬀerential equations, just the initial conditions. • Which ﬂeet will prevail in Skirmish “A”? How many ships will remain in that ﬂeet? (4 points) • Which ﬂeet will prevail in Skirmish “B”? How many ships will remain in that ﬂeet? (4 points) • Did the French-Spanish commandants use Admiral Nelson’s strategy successfully? You shouldn’t have to do any more calculations for this, just use the ﬁgures from parts (a) and (b). (Remember that partial ships can’t ﬂoat and won’t be able to ﬁght.) (2 points) 1 3. Evaluate the integral x3 1 − x4 dx. (10 points) 0 n2 4. Determine if the sequence an = is convergent or divergent. (10 points) 1+n 5. Evaluate the integral x cos x dx. (10 points) 4 3 y2 1 0 0 0.5 1 1.5 2 2.5 3 x Exercises #6 and #7 refer to the graph above. 6. Find the area of the region bounded by the curves y = e−x + 2, y = x − 1 , x = 1, and x = 2. 2 (12 points) 7. Find the volume of the solid obtained by rotating the region bounded by the curves y = e−x + 2, y = x − 1 , x = 1, and x = 2 about the x-axis. 2 (12 points) x 8. Let g(x) = f (t) dt, where f is the function whose graph is shown. Answer the following: 0 (3 points for each part, 12 points total) y (a) Evaluate g(0), g(1), and g(2). f (b) Estimate g(3), g(4), and g(5). 1 (c) On what interval is g increasing? 1 t (d) Sketch a rough graph of g. ∞ xn 9. Find the interval of convergence of the series . (10 points) n=1 n2n > with(student) : > # Exercise 1 f := x -> -1*x^2 + 4*x - 2 ; 2 f := x -x + 4 x - 2 > leftbox( f(x) , x = 0..4 , 4 ) ; > Int( -1*x^2 + 4*x - 2 , x = 0..4 ) = int( -1*x^2 + 4*x - 2 , x = 0..4 ) ; 4 2 8 -x + 4 x - 2 dx = 3 0 > Int( -1*x^2 + 4*x - 2 , x ) = int( -1*x^2 + 4*x - 2 , x ) + C ; 2 1 3 2 -x + 4 x - 2 dx = - x + 2 x - 2 x + C 3 > > # Exercise 2, part (a) x0 := 11 ; y0 := 12 ; C := y0^2 - x0^2 ; y^2 - x^2 = C ; y^2 - 0^2 = C ; y = C^(1/2.) ; x0 := 11 y0 := 12 C := 23 2 2 y - x = 23 2 y = 23 y = 4.795831523 > # Exercise 2, part (b) x0 := 11 ; y0 := 9 ; C := y0^2 - x0^2 ; y^2 - x^2 = C ; 0^2 - x^2 = C ; x = (-C)^(1/2.) ; x0 := 11 y0 := 9 C := -40 2 2 y - x = -40 2 -x = -40 x = 6.324555320 > # Exercise 2, part (c) x0 := 6 ; y0 := 4 ; C := y0^2 - x0^2 ; y^2 - x^2 = C ; 0^2 - x^2 = C ; x = (-C)^(1/2.) ; x0 := 6 y0 := 4 C := -20 2 2 y - x = -20 2 -x = -20 x = 4.472135955 > > with(DEtools) : > # Exercise 2, part (a) phaseportrait( [diff(x(t),t)=-y(t), diff(y(t),t)=-x(t)], [x(t),y(t)], t = 0..10, [[x(0)=11,y(0)=12]], stepsize=1, x = 0..30, y = 0..30 ) ; > > phaseportrait( [diff(x(t),t)=-y(t), diff(y(t),t)=-x(t)], [x(t),y(t)], t = 0..10, [[x(0)=11,y(0)=9]], stepsize=1, x = 0..30, y = 0..30 ) ; > restart ; # this clears out the value of 'C' which we use again below > # Exercise 3 Int( x^3*(1-x^4)^(1/2) , x = 0..1 ) = int( x^3*(1-x^4)^(1/2) , x = 0..1 ) ; 1 3 4 1 x 1 - x dx = 6 0 > Int( x^3*(1-x^4)^(1/2) , x ) = int( x^3*(1-x^4)^(1/2) , x ) + C ; (3/ 2) 3 4 1 4 x 1 - x dx = - (1 - x ) +C 6 > # Exercise 4 Limit( n^2/(1+n) , n = infinity ) = limit( n^2/(1+n) , n = infinity ) ; 2 n lim = n 1+n > # Exercise 5 Int( x*cos(x) , x ) = int( x*cos(x) , x ) + C ; x cos(x) dx = cos(x) + x sin(x) + C > > # the following code is how we created the figure for Exercises 6 and 7 > with(plots) : > p1 := plot( exp(-x) + 2 , x = 0..3 , y = 0..4 , thickness = 2 , color = blue ) : > p2 := plot( x - 1/2, x = 0..3 , y = 0..4 , thickness = 2 , color = green ) : > p3 := implicitplot( x = 1 , x = 0..3 , y = 0..4 , thickness = 2 , color = red ) : > p4 := implicitplot( x = 2 , x = 0..3 , y = 0..4 , thickness = 2 , color = red ) : > plots[display]({p1,p2,p3,p4}) ; > > # Exercise 6 Int( exp(-x)+2 - (x - 1/2) , x = 1..2 ) = int( exp(-x)+2 - (x - 1/2) , x = 1..2 ) ; 2 (-x) 5 (-1) (-2) e + - x dx = e -e +1 2 1 > int( exp(-x)+2 - (x - 1/2) , x = 1..2.0 ) ; 1.232544158 > # Exercise 7 Int( pi*( (exp(-x)+2)^2 - (x - 1/2)^2 ) , x = 1..2 ) = int( pi*( (exp(-x)+2)^2 - (x - 1/2)^2 ) , x = 1..2 ) ; 2 2 2 (-x) 1 (-2) (-1) (-4) 7 35 1 e +2 - x- dx = - e +4 e + - e 2 2 12 2 1 > int( pi*( (exp(-x)+2)^2 - (x - 1/2)^2 ) , x = 1..2.0 ) ; 3.905353121 > > # Exercise 9 Sum( x^n/(n*2^n) , n = 1..infinity ) = sum( x^n/(n*2^n) , n = 1..infinity ) ; n x 1 = -ln 1 - x n= 1 n 2 n2 > # think about what happens when 'x < 2', 'x > 2' , or 'x = 2' > # Ratio Test Limit( abs((x^(n+1)/((n+1)*2^(n+1)))/(x^n/(n*2^n))) , n = infinity ) ; (n + 1) n x n2 lim n (n + 1) n ( n + 1) 2 x > # can you see how things cancel out in the expression above? limit( (x^(n+1)/((n+1)*2^(n+1)))/(x^n/(n*2^n)) , n = infinity ) ; # be careful with this simplified version -- I dropped the absolute value here! 1 x 2 >