Math 132 Fall 2005 Exam I 1 . A Riemann sum for a function on an interval is said to be a n upper Riemann sum if, for each , the point in the subinterval is chosen so that ismaximized. Calculate the upper Riemann sum for , , and . (Use a partition of [a,b] into equal length subintervals.)a) 14 b) 16 c) 18 d) 20 e) 22 f) 24 g) 26 h) 28 i) 30 j) 32Solution : ( i )The nodes (points of the uniform partition) are> a := -1; b := 3; N := 4; Delta := (b-a)/N;> for j from 0 to N dox[j] := a + j*Delta;od;> f := x -> x^3-3*x+2;> Diff('f(x)',x)=factor(D(f)(x)); This calculation tells us that decreases from -1 to 1 ( = ( - )( + ) so ) and increases for > 1 . Therefore:> xi[1] := x[0]; xi[2] := x[1]; xi[3] := x[3]; xi[4] := x[4];> fnGraph := plot(f(x),x=-1..3, color=PLUM, thickness=2): > nodes := plot( [ seq([xi[j],f(xi[j])], j = 1..N)], style = POINT, symbol = CIRCLE, color = NAVY):> plots[display](fnGraph, nodes);> rect[1] := plottools[rectangle]([x[0],f(x[0])], [x[1],0], color = wheat):rect[2] := plottools[rectangle]([x[1],f(x[1])], [x[2],0], color = wheat):rect[3] := plottools[rectangle]([x[2],f(x[3])], [x[3],0], color = wheat):rect[4] := plottools[rectangle]([x[3],f(x[4])], [x[4],0], color = wheat):> plots[display](fnGraph, nodes,seq(rect[j],j=1..N));> ANSWER := (f(xi[1])+f(xi[2])+f(xi[3])+f(xi[4]))*Delta;2. Calculate .a) 1 b) c) d) 2 e) f) g) h) i) j) Solution : ( a )> F := unapply(int(sec(theta)*tan(theta), theta), theta);> F(Pi/3) - F(0);3. An antiderivative of is the function . If , what is ?a) b) c) d) e) f) g) 1 h) i) 2 j) 2 Solution : ( i )> restart;> F := x -> (x+exp(x))/(1+exp(x));> eqn1 := Int(f(x)+c,x = 0 .. 1) = F(1) - F(0) + c;> eqn2 := 5/2 = rhs(eqn1);> solve(eqn2,c);4. Calculate . a) 1 b) 2 c) d) e) f) g) h) i) j) Solution : ( c )> F := x -> x*ln(x); #This is an antiderivative of the integrand > F(exp(1)) - F(1);5. Suppose that and . What is ?a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10 Solution : ( d )> restart;> eqn1 := int(x^2+f(x),x = 0..3) = 17; > eqn2 := int(x^2,x = 0..3) + int(f(x),x = 0..3)= 17;> eqn3 := 9 + int(f(x),x = 0 .. 2) + int(f(x),x = 2 .. 3) = 17;> eqn4 := 9 + 4 + int(f(x),x = 2 .. 3) = 17;> solve( eqn4, int(f(x),x = 2 .. 3));6. Suppose that . The Mean Value Theorem for Integrals asserts that there is a pointin the interval [-1,2] such that where is the average value of for in the interval [1,7]. What is ?a) b) c) d) e) f) g) h) i) j) Solution : ( f )> f := x -> x^2+6*x; interval := -1 .. 2;> Ave := int(f(x), x=interval)/3;> solve(f(x) = Ave, x); 7. Calculate at .a) 0 b) 1 c) 2 d) 3 e) 4 f) 5 g) 6 h) 7 i) 8 j) 9 Solution : ( c )> J := Int((9*t+tan(Pi*t/4))/(t^2+4),t = -1 .. x);> Integrand := student[integrand](J);> simplify(subs(t=1, Integrand)); 8. Suppose that . What is ? (The derivative of F(x) at ). a) 1 b) c) d) 2 e) f) g) h) i) j) Solution : ( b ) > restart: with(student):> F := (x) -> Int(sqrt(7/2+t^2),t = 0..sin(x));> D(F)(x); > simplify(D(F)(Pi/4));Where answer comes from:> derivative := subs(t=sin(x), integrand(F(x)))*D(sin)(x);> simplify(subs(x=Pi/4, derivative)); 9 . Suppose that . What is D(F)(2)? (The derivative of F(x) at x =5). a) 13 b) 20 c) 27 d) 33 e) 40 f) 47 g) 53 h) 60 i) 67 j) 73 Solution : ( f )> F := (x) -> Int(sqrt(144+t^2),t = x .. (3*x+1));> simplify(D(F)(5));Where answer comes from:> h := x -> 3*x+1; g := x -> x;> derivative := subs(t=h(x), integrand(F(x)))*D(h)(x) - subs(t=g(x), integrand(F(x)))*D(g)(x);> simplify(subs(x=5, derivative)); 10. Calculate . a) 3 b) 4 c) 6 d) 8 e) 9 f) 8 g) 12 h) 15 i) 16 j) 20Solution : ( h )> J1 := Int(4*cos(x)*(sin(x)+1)^3,x = 0 .. Pi/2);> J2 := changevar(u = sin(x) + 1, J1, u);> value(J2); 11. Calculate .a) b) c) d) e) f) g) h) i) j) Solution : ( i )> J1 := Int(x*sqrt(x-3),x = 3..4);> J2 := changevar(u = x-3, J1, u);> J3 := Int(expand(integrand(J2)), u = 0 .. 1);> value(J3);12. Calculate .a) b) c) d) e) f) g) h) i) j) Solution : ( e )> J1 := Int(1/(x+x*ln(x)),x = 1 .. exp(1));> J2 := Int(factor(integrand(J1)), x = 1 .. exp(1));> J3 := changevar(u=1+ln(x), J2, u);> value(J3);13. Find the solutions x = a and x = b of the equation sin(x) = cos(x) in the first and third quadrantsrespectively. Calculate the area between and for in [a,b].a) 1 b) c) 2 d) e) f) g) h) i) j) 4 Solution : ( e )> solve( sin(x) = cos(x) );> plot([sin(x) , cos(x)], x = Pi/4 .. 5*Pi/4, color = [NAVY,MAROON]);> int(sin(x) - cos(x), x = Pi/4 .. 5*Pi/4);> 14. At irregular intervals during the first 10 seconds of a race, a radar gun recorded the following speeds (in m/s) of a runner: .Given that distance at time is expressed by the formula , estimate thedistance (in m) the runner has covered during those 10 seconds.
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