Practice test. The real test is Monday, February 25, 2008 1. Find the multiplicative constants A and B so that -1 2 x 3 F(x) = A Tan ( x ) + B (x + e ) x x 2 x F'(x) = ------- + (x + e ) (1+e ) 4 1+x Solution: 2 x x 2 x F'(x) = A ----------- + B 3(x+e ) ( 1 + e ) 2 2 1+( x ) Answer A = 1/2 and B = 1/3 2. Find where the curves f(x) = 10/x and g(x) = 7-x intersect. Then find the area of the bounded region determined. Solution: 10/x = 7-x 10 = 7x-x^2 x^2 -7x +10 = 0 (x-2)(x-5) x=2 x=5 x=5 INT 7-x - 10/x dx x=2 | | x=5 | 7x - x^2/2 -10 ln(x) | = 35 -25/2 -10 ln(5) -(14-2-10ln(2) | | x=2 21 answer = -- + 10 ln(2) - 10 ln[5] = 1.33709 2 ------------------------------------------------------------------------- a = Plot[{7-x,10/x},{x,1,6}]; b = Show[a,PlotLabel->"y = 10/x and y = 7-x Practice Test Question 2", AspectRatio->Automatic]; Display["p2.ps",b]; Integrate[7-x-10/x,{x,2,5}]; a = Plot[{7-x,10/x},{x,2,5}]; h = Table[Plot[n/5,{x,2,5},PlotStyle->{RGBColor[1,0,0]}],{n,10,25}]; v = Table[ParametricPlot[{n/5,y},{y,2,5},PlotStyle->{RGBColor[1,0,0]}],{n,10,25}]; b = Show[a,h,v,PlotLabel->"y=10/x; y = 7-x 2, (1/5)x(1/5) grid", AspectRatio->Automatic]; Display["p2x.ps",b]; ------------------------------------------------------------------------- 4 3 3. Find d/dx INT ln(t ) dt Sin[x] Solution: 3 y' = ln( Sin [x] ) (-Cos[x]) ---------------------------------------------------- a = Plot[ Log[t^3],{t,0.1, 10}]; b = Show[a,PlotLabel->"P3, Log[t^3]",AspectRatio->Automatic, PlotRange->All]; Display["p3.ps",b]; f[x_] := NIntegrate[ Log[t^3],{t,Sin[x],4}]; c = Plot[ f[x],{x,0,Pi} ]; d = Plot[ Log[ Sin[x]^3 ] (-Cos[x] ),{x,Pi/20,19 Pi/20},PlotStyle->{RGBColor[1,0,0]}]; e = Show[c,d,PlotLabel->"P3 black = curve;red=derivative",AspectRatio->Automatic, PlotRange->All]; Display["p3x.ps",e]; -------------------------------------------------------------------- (2/x) 4. Find Limit (1 + Sin[x]) x->0 Solution: (2/x) y = (1 + Sin[x]) ln(y) = 2/x ln(1+Sin[x]) ln(1+Sin[x]) ln(y) = 2 -------------- x Cos[x] ----------- 1+Sin[x] Limit ln(y) = 2 Limit ---------- = 2 x->0 x->0 1 2 Answer: Limit y = e = 7.38906 x->0 ------------------------------------------------------- a = Plot[(1+Sin[x])^(2/x),{x,-0.5,10}]; b = ListPlot[{{0,E^2}},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P4; y = (1+Sin[x])^(2/x)",PlotRange->All]; Display["p4.ps",c]; ------------------------------------------------------------ 2 x 5. Use substitution to evaluate INT --------- dx 6 1 + x Solution: u = x^3 du = 3 x^2 dx du/3 INT ------------ 1+u^2 1/3 ArcTan[u] + C Answer: 1/3 ArcTan[x^3] + C x=1 6. INT ArcTan[x] dx x=0 Solution: ArcTan[x] x dx --------- dx 1 + x^2 x dx = x ArcTan[x] - INT ---------- 1+x^2 | | x=1 = | x Arctan[x] - 1/2 ln(1+x^2) | | | x=0 Answer: Pi/4 -1/2 ln(2) 3 4 7. Find INT Sin [t] Cos [t] dt 2 4 Solution: INT Sin [t] Cos [t] Sin[t] dt 2 4 INT (1-Cos [t]) Cos [t] Sin[t] dt 4 6 INT (Cos [t] - Cos [t]) Sin[t] dt 5 7 -Cos [t] Cos [t] ------- + --------- + C 5 7 4 8. The region bounded by y = Sqrt[x] and y = x is rotated about the y-axis. Find the volume of the solid generated. x=1 Volume = INT 2 Pi x (Sqrt[x] - x^4) dx x=0 3/2 5 = 2 Pi INT x - x dx | 5/2 6 | x=1 | x x | = 2 Pi | ----- - ----- | | 5/2 6 | x=0 = 2 Pi (2/5 - 1/6) = 7 Pi/15 ------------------------------------------ a = Plot[{Sqrt[x],x^4},{x,0,1.5},AspectRatio->Automatic]; b = Show[a,PlotLabel->"y = Sqrt[x], y = x^4 Practice Test Question 8"]; Display["p8.ps",b]; a = ParametricPlot3D[ { x Cos[t], x Sin[t], Sqrt[x]},{t, Pi/8,6 Pi/4},{x,0,1}]; b = ParametricPlot3D[ { x Cos[t], x Sin[t], x^4},{t, Pi/8,6 Pi/4},{x,0,1}]; c = Show[a,b,PlotLabel->"Practice Test Question 9"]; Display["p8x.ps",c]; ------------------------------------------ 9. The signum function sgn(x) is defined by sgn(x) = 1 for x>0, sgn(0) = 0, Sgn(x) = -1 for x < 0. Let a < 0 and b > 0. b Find INT sgn(x) dx in terms of a and b. Justify your answer. a Solution: | (------------------------------- | Area = b | a | | -------|----------------*----------------|------------------- | Area = a | b | | ----------------------) | Answer: a+b. Actually the area to the left is negative, but so in a so the area to the left does equal a. x 10. The region determined by y = e, y=x, and x=0, x=2 is rotated about the x-axis. Set up, DO NOT EVALUATE the definite integral for this volume. Solution: x=2 x 2 2 INT Pi (e ) - Pi x dx x=0 x=2 2x 2 Pi INT e - x dx x=0 | 2x 3 | x=2 Pi |1/2 e - 1/3 x | | | x=0 Pi ( 1/2 e^4 - 8/3 - (1/2)) = ( 1/2 e^4 - 19/6 ) Pi = 75.8142 -------------------------------------------------------------------------------- Integrate[Pi (E^x)^2 - Pi x^2,{x,0,2}]; a = Plot[E^x,{x,0,2}]; b = Plot[x,{x,0,2}]; c = ParametricPlot[{2,y},{y,0,E^2}]; d = ParametricPlot[{0,y},{y,0,E^2}]; e = Show[a,b,c,d,AspectRatio->Automatic, PlotLabel->"p10; y=e^x, y = x "]; Display["p10.ps",e]; a = ParametricPlot3D[ { x, E^x Cos[t], E^x Sin[t] },{t, -Pi/2, Pi/2},{x,0,2}]; b = ParametricPlot3D[ { x, x Cos[t], x Sin[t] },{t, -Pi/2, Pi/2},{x,0,2}]; c = ParametricPlot3D[ { 2, y Cos[t], y Sin[t] },{t, -Pi/2, Pi/2},{y,2,E^2}]; d = ParametricPlot3D[ { 0, y Cos[t], y Sin[t] },{t, -Pi/2, Pi/2},{y,0,1 }]; e = Show[a,b,c,d,PlotLabel->"P10 y=e^x, y=x [0,2] about X-axis"]; Display["p10x.ps",e]; -------------------------------------------------------------------------------- dx 11. Evaluate INT ---------------- using triangle substitution. 2 2 x Sqrt[x - 9 ] Draw and label the triangle. Solution: /| / | / | / | x / | Sqrt[x^2-9] / | / | / | / t | /_________| 3 x = 3 Sec[t] Sqrt[x^2-9] = 3 Tan[t] dx = 3 Sec[t] Tan[t] dt 3 Sec[t] Tan[t] dt INT ------------------------- 2 9 Sec [t] 3 Tan[t] dt 1/9 INT -------------------- Sec[t] 1/9 INT Cos[t] 1/9 Sin[t] Sqrt[x^2 - 9] Answer = 1/9 ---------------- x x=+Infinity dx 12. Find INT ------------ x=-Infinity 2 1+x Solution: | x = + Infinity ArcTan[x] | | x = - Infinity Answer = Pi/2 - (-Pi/2) = Pi x=2Pi Cos[x] 13. Find INT -------- dx x=0 1+Sin[x] Solution: The integral is improper because at x = 3 Pi/2 the denominator is zero. x=b Cos[x] | x=b INT ------------ dx = ln(1+Sin[x]) | x=0 1+Sin[x] | x=0 = ln(1+Sin[b]) as b approaches 3 Pi/2, Sin[b] approaches -1 and the ln(1+Sin[b]) approaches - Infinity. Therefore the integral does not exist. ---------------------------------------------------- a = Plot[Cos[x]/(1+Sin[x]),{x,0,3 Pi/2-0.01}]; b = Plot[Cos[x]/(1+Sin[x]),{x,3 Pi/2+0.01,2 Pi}]; c = ParametricPlot[{3 Pi/2,y},{y,-120,120},PlotStyle->{RGBColor[1,0,0]}]; d = Show[a,b,c,PlotLabel->"p18 Cos[x]/(1+Sin[x])"]; Display["p13.ps",d]; ------------------------------------------------ x=4 dx 14. INT -------------- x=2 Sqrt[4x-x^2] Solution: x=4 dx INT -------------------- x=2 Sqrt[ 4 - (x-2)^2 ] x=4 dx 1/2 INT -------------------- x=2 Sqrt[ 1 - (( x-2)/2 )^2 ] x=4 1/2 dx INT -------------------- x=2 Sqrt[ 1 - (( x-2)/2 )^2 ] | x=4 ArcSin[ (x-2)/2 ] | | x=2 ArcSin[1] - ArcSin[ 0] Answer: Pi/2 - (0) = Pi/2 Integrate[1/Sqrt[4x-x^2],{x,2,4}]; 15. Let R be the region in the first quadrant below the curve -2/3 y = x and to the left of x=1. (a) Find the area of R. (b) Find the volume of the solid generated by revolving R about the x-axis. Solution: x=1 -2/3 (a) INT x dx x=0 1/3 | x = 1 x | ---------- | = 3 1/3 | x = 0 x=1 -2/3 2 (b) INT Pi (x ) dx x=0 -4/3 Pi INT x dx | -1/3 | x=1 | x | Pi| ------- | | -1/3 | x=b Pi ( -3 + 3 b^(-2/3) ) = + INFINITY as b approaches 0. -------------------------------------------------------------------------------- a = Plot[ x^(-2/3),{x,0.01, 1}]; b = Show[a, PlotLabel->"p15; y= x^(-2/3)"]; Display["p15.ps",b]; c = ParametricPlot3D[ { x,x^(-2/3) Cos[t], x^(-2/3) Sin[t] },{t,-Pi/2,Pi/2},{x,0,1}]; d = ParametricPlot3D[ { 1, y Cos[t], y Sin[t] },{t, -Pi/2, Pi/2},{y,0,1}]; e = ParametricPlot3D[ { 0, y Cos[t], y Sin[t] },{t, -Pi/2, Pi/2},{y,0,10}]; f = Show[c,d,e,PlotLabel->"P15 y = x^(-2/3) about X-axis",PlotRange->All]; Display["p15x.ps",f]; -------------------------------------------------------------------------------- 16. Show that if p(x) is any polynomial, then p(x) Lim ------------ = 0 x->Infinity x e Solution: Using L'Hopital's rule, we can differentiate the numerator until it becomes a constant. The denominator will remain the same. constant Eventually we have Limit ---------- = 0 x->Infinity x e 17. Find Limit 2 x Ctn[x] x->0 Solution: 2x 2 Limit 2 x Ctn[x] = Limit --------- = Limit ------- = 2 x->0 x->0 Tan[x] x->0 2 Sec [x] -------------------------------------------------------- a = Plot[2 x Cot[x] ,{x,-Pi/4,Pi/4}]; b = Show[a,PlotLabel->"Practice Test Question 17"]; Display["p17.ps",b]; -------------------------------------------------------- Tan[3x] 18. Find Limit -------- x-> Pi/2 Tan[x] Solution: 2 Tan[3x] 3 Sec [3x] Limit ---------- = Limit ------------ x-> Pi/2 Tan[x] x-> Pi/2 2 Sec [x] 2 3 Cos [x] -6 Cos[x] Sin[x] Limit ------------------- = Limit ----------------- x-> Pi/2 2 x-> Pi/2 -6 Cos[3x] Sin[3x] Cos [3x] 2 2 -Sin [x] + Cos [x] 1 Limit -------------------------- = ----------- x-> Pi/2 2 2 3 -3 Sin [3x] + 3 Cos [3x] Answer: 1/3 ------------------------------------------------------------------- a = Plot[Tan[3x]/Tan[x],{x,Pi/4,3Pi/4}]; b = ListPlot[{{Pi/2,1/3}},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P18: y=Tan[3x]/Tan[x]"]; Display["p18.ps",c]; -------------------------------------------------------------------