Math 132 Fall 2007 Exam I Notation: D(f)(x) means: "the derivative of f evaluated at x." For example, if = = = = ( )f x x3, then = = = = ( )( )D f x 3 x2 and = = = = ( )( )D f 5 75. 1. A Riemann sum ∑∑∑∑ = = = = j 1N( )f sj∆∆∆∆ x for a function f on an interval [ ],a b is said to be an upper Riemann sum if, for each j, the point sj in the j th subinterval is chosen so that ( )f sj is the maximum value of ( )f x for x in the j th subinterval. Using the uniform partition of = = = = [ ],a b [ ],0 6 into 3 subintervals, calculate the upper Riemann sum for the function f whose graph is given.a) 13 b) 14 c) 15 d) 16 e) 17 f) 18 g) 19 h) 20 i) 21 j) 22 Solution: (h)> Delta := (6-0)/3; upperRiemannSum := f(0.5)*Delta + f(2)*Delta + f(2)*Delta; upperRiemannSum := subs({f(0.5)=4, f(2)=3}, upperRiemannSum); := ∆ 2 := upperRiemannSum + 2 ( )f 0.5 4 ( )f 2 := upperRiemannSum 202. The expression is a right Riemann sum. What is lim → → → → N ∞∞∞∞SN ? a) 16 b) 20 c) 24 d) 32 e) 36 f) 44 g) 56 h) 60 i) 64 j) 72Solution: ( h )A right Riemann sum of ( )f x on the interval [a,b] has the form > restart:> f(a+(b-a)/N)*(b-a)/N+f(a+2*(b-a)/N)*(b-a)/N+`...`+f(a+N*(b-a)/N)*(b-a)/N; + + + f + a − b aN( ) − b aNf + a2 ( ) − b aN( ) − b aN...( )f b ( ) − b aNComparing this with the given sum we see that> a := 2; Delta := 2/N; b := a + N*Delta; f := x -> x^3; := a 2 := ∆2N := b 4 := f → x x3Since the limit of the Riemann sums is the Riemann integral of f over [a,b], we have:> Answer := Int(f(x), x = a .. b); answer = value(Answer); := Answer d⌠⌡24x3x = answer 603. Calculate d⌠⌠⌠⌠⌡⌡⌡⌡0ππππ3( )sec θθθθ2θθθθ.a) 1 b) 2 c) 3 d) − − − − 2 1 e) − − − − 3 1f) 32 g) 22 h) 2 3 i) 3 2 j) 3 3 Solution: ( c ) > indefiniteIntegral := Int(sec(theta)^2, theta); := indefiniteIntegral d⌠⌡( )sec θ2θ> indefiniteIntegralEvaluated := value( indefiniteIntegral ); := indefiniteIntegralEvaluated( )sin θ( )cos θ> answer := subs(theta=Pi/3, indefiniteIntegralEvaluated)-subs(theta=0, indefiniteIntegralEvaluated); := answer − sinπ3cosπ3( )sin 0( )cos 0> simplifiedAnswer := simplify(answer); := simplifiedAnswer 34. Suppose that = = = = ( )f x + + + + + + + + 4 x24 x 20 + + + + x2x . What is d⌠⌠⌠⌠⌡⌡⌡⌡14( )( )D f x x ?a) - 9 b) - 7 c) - 5 d) - 3 e) - 1 f) 1 g) 3 h) 5 i) 7 j) 9 Solution: (a) > restart;> f := x -> (4*x^2+4*x+20)/(x^2+x);:= f → x + + 4 x24 x 20 + x2x> answer := f(4) - f(1); #Fundamental Theorem of Calculus, Part I := answer -95. Suppose that = = = = d⌠⌠⌠⌠⌡⌡⌡⌡142 ( )f x x 5 , = = = = d⌠⌠⌠⌠⌡⌡⌡⌡−−−−11( )f x x92 , and = = = = d⌠⌠⌠⌠⌡⌡⌡⌡−−−−14 + + + + ( c ( )f x ) x 22. What is c ?a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10 Solution: ( c ) > restart;> := eqn1 = d⌠⌡−14 + c ( )f x x 22 := eqn1 = d⌠⌡-14 + c ( )f x x 22> eqn2 := c = solve(eqn1, c); := eqn2 = c − + 15d⌠⌡-14( )f x x225> eqn3 := subs(int(f(x),x = -1 .. 4) = int(f(x),x = -1 .. 1)+(1/2)*int(2*f(x),x = 1 .. 4), eqn2); := eqn3 = c − − + 15d⌠⌡-11( )f x x110d⌠⌡142 ( )f x x225> subs ,{ }, = d⌠⌡−11( )f x x92 = d⌠⌡142 ( )f x x 5 eqn3 = c 36. Suppose that = = = = ( )f x − − − − 3 x22 x . The Mean Value Theorem for Integrals asserts that there is a point c in the interval [1,4] such that = = = = ( )f c fave where fave is the average value of ( )f x for x in the interval [1,4]. What is c? a) 4/3 b) 5/3 c) 2 d) 7/3 e) 8/3 f) 3 g) 10/3 h) 11/3 i) 5/2 j) 7/2Solution: ( e ) > a := 1; b := 4; f := x -> 3*x^2-2*x; := a 1 := b 4 := f → x − 3 x22 x> m := Int(f(x),x=1..4)/(b-a); := m13d⌠⌡14 − 3 x22 x x> m := value(m); := m 16> solve(f(c) = m, c);,83-2Only the first solution is in the interval.7. Calculate d d x d⌠⌠⌠⌠⌡⌡⌡⌡0x + + + + 7 t 22 + + + + t34t at = = = = x 2. a) 0 b) 1 c) 2 d) 3 e) 4 f) 6 g) 9 h) 12 i) 16 j) 18 Solution: ( d ) > f := t -> (7*t+22)/(t^3+4); := f → t + 7 t 22 + t34> Answer := subs(t=2, f(t)); := Answer 38. Suppose that = = = = ( )F x d⌠⌠⌠⌠⌡⌡⌡⌡0 x2 + + + + t220 t. What is ( )( )D F 2 ? a) 16 b) 20 c) 24 d) 32 e) 36 f) 44 g) 48 h) 60 i) 64 j) 72 Solution: ( c ) > F := (x) -> Int(sqrt(t^2+20),t = 0 .. x^2); := F → x d⌠⌡0x2 + t220 t> D(F)(x);2 x + x420> D(F)(2);4 36> simplify( D(F)(2) );24 9. Suppose that = = = = ( )F x d⌠⌠⌠⌠⌡⌡⌡⌡ ( )cos x ( )sin x + + + + 12t2t. What is ( )D Fππππ4 ? a) 0 b) 1/2 c) 1 d) 2 e) 3 f) 4 g) 2 h) 2 2 i) 22 j) 3 2 Solution: ( g ) > with(student):> v := x -> sin(x); u := x -> cos(x); F := (x) -> Int(sqrt(1/2+t^2), t = u(x) .. v(x)); := v sin := u cos := F → x d⌠⌡( )u x( )v x + 12t2t>
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