Review of Integral Calculus
Copyright 1998
Department of Mathematics
University of Georgia
Athens, Georgia
Carol W
. Penney
As you approach the end of your course in integral calculus, you want to review the course to put into perspective what you have learned and to refresh your memory of topics you may not have thought about in a while. This project is a review of some of the main topics of the course, using Maple to check your answers and to illustrate some of the ideas you have studied. These ten topics constitute a framework of the course, but, of course, do not cover all aspects of the course, so do not count on this project to adequately prepare you for the final examination in the course. In particular, reasoning is required to set up the solutions of word problems, and Maple will not do the reasoning for you; this review will not be very useful for studying word problems, which are in many ways the heart of the course. This review is most useful for practicing routine calculus exercises.
How should you use this review? You should work each problem with pencil and paper, then use the Maple commands to check your work. If you "cheat" by looking entering Maple's commands before working the problems for yourself, you will destroy the value of this review, much as reading the last chapter of a mystery destroys its value.
Definition of the Integral
This first course in calculus is a study of the definite integral. You certainly should be able to state the definition of the definite integral of a function f(x) from x=a to x=b and use this definition to evaluate the definite integral of a function.
| > | restart: |
Problem:
Consider the function f(x)=x^2 on the closed interval [1,5]. Use the
definition of the definite integral
to evaluate
. Do this in the following steps:
Define a Riemann sum, using n subintervals. First define
to be a function of n; we will call this function deltax. You will have to refer to this function as deltax(n) in future calculations.
| > | f:=x->x^2; |
| > | deltax:=n-> |
| > |
Define
, which we here call x(i,n), since it will depend upon both i and n, in terms of 1 and i and deltax(n). You will have to refer to this function as x(i,n) in future calculations.
| > | x:=(i,n)-> |
Define xstar (your choice, preferably the right-hand endpoint or the left-hand endpoint) in terms of the function x(i,n) you defined in the previous command. It is a function of both i and n, so you will have to refer to it in future calculations as xstar(i,n).
| > | xstar:=(i,n)-> |
Now define the Riemann sum as a function of n, using xstar(i,n) and deltax(n):
| > | riemannsum:=n-> |
Now sketch a diagram of a Riemann sum, using 15 subintervals, to make sure that you have defined your Riemann sum correctly. First define n to be 15, then define the Riemann rectangles to be displayed.
| > | n:=15: |
| > | riemannrectangle:=(i,n)->rectangle([x(i-1,n),0], [x(i,n),f(xstar(i,n))],color=red); |
Form the sequence of all of these rectangles,
| > | rectangles:=[seq(riemannrectangle(i,n),i=1..n)]: |
and, since you want to simultaneously display the graph of the function and the Riemann rectangles, open up the packages plots and plottools:
| > | with(plots):with(plottools): |
and display the graph of f along with the rectangles.
| > | display(plot(f(x),x=0..5,title=`Riemann Sum`,color=navy,thickness=2), rectangles); |
Look at the diagram to determine whether or not you have correctly defined the Riemann sum.
| > |
Remove (or unassign ) the value of n, so that you can calculate the limit of the nth Riemann sum:
| > | unassign('n'); |
Now simplify, by hand, the Riemann sum as a function of n, then find its limit. Compare with the following:
| > | simplify(riemannsum(n)); |
| > | limit(simplify(riemannsum(n)),n=infinity); |
| > | int(f(x),x=1..5); |
| > |
You can, of course, practice with various functions on intervals of your choice.
Fundamental Theorem of Calculus
State the Fundamental Theorem of Calculus. Look up this theorem in the text to make sure that you have stated it correctly. Then work the following:
Problem:
Suppose that
. Find
.
Solution:
Suggestion: let
. Find dy/dx. Then check your answer with Maple:
| > | y:=x->int(sqrt(t^3+1),t = 0 .. x^2); |
| > | D(%)(x); |
Problem:
Suppose that
. Find
.
| > | y:=x->int(sin(t^2),t = 0 .. exp(3*x)); |
| > | D(%)(x); |
| > |
Area
Problem: Find the intersections of the graphs of f(x)=x^2-x and g(x)=1-x^3, sketch the lamina bounded by these graphs, then find the area of this lamina.
Solution:
| > | restart: |
| > | f:=x->x^2-x; |
| > | g:=x->1-x^3; |
Find intersections, then enter appropriate values of a and b in the following command.
| > | plot([f(x),g(x)],x=a..b,linestyle=[1,3]); |
Find the appropriate integrand and the appropriate values of c and d, and compute the integral.
| > | int( ); |
| > |
Problem: Find the area of the region bounded by the graphs of f(x)=sin(x) and g(x)=x^2.
Solution:
| > | restart: |
| > | f:=x->sin(x); |
| > | g:=x->x^2; |
Find the intersection points of these curves:
| > | fsolve( ); |
| > | plot([f(x),g(x)],x=-3..3,linestyle=[0,3]); |
Now determine the area of this region.
| > | int(); |
| > |
Volume by the Method of Slices
Problem: Sketch the lamina bounded by the graph y=sin(x) and y=0, from x=0 to x=Pi, then find the volume of the solid generated by revolving this curve about the x-axis. Be sure to set this up and antidifferentiate by hand, then compare with this:
Solution:
| > | plot([sin(x),0],x=0..Pi,thickness=3,color=red); |
| > | int(Pi*sin(x)^2,x=0..Pi); |
Problem: Revolve the region bounded by y=exp(x), y=0, x=0, and x=1 about the x-axis. Be sure to set this up and antidifferentiate by hand, then compare with:
Solution:
| > | int(Pi*exp(2*x),x=0..1); |
| > |
Problem: Revolve this same region, the region bounded by y=exp(x), y=0, x=0, and x=1, about the line y= -1. Again, work by hand first, then check with:
Solution:
| > | int(Pi*((exp(x)+1)^2-1),x=0..1); |
| > |
Volume by the Method of Shells
Problem: Revolve the region bounded by y=sin(x), y=0, from x=0 to x=Pi about the y-axis to generate a solid. Find the volume of this solid. Be sure to set this up and antidifferentiate by hand, then compare with:
Solution:
| > | int(2*Pi*x*sin(x),x=0..Pi); |
Revolve this same region about the line x=Pi. What do you expect to get?
| > | int(2*Pi*(Pi-x)*sin(x),x=0..Pi); |
| > |
Problem: Revolve the region bounded by y=exp(x), y=0, x=0, and x=1 about the y-axis to generate a solid. Find the volume of this solid. Be sure to set this up and antidifferentiate by hand, then compare with:
Solution:
| > | int(2*Pi*x*exp(x),x=0..1); |
Problem: Revolve this same region, the region bounded by y=exp(x), y=0, x=0, and x=1, about the line x=1. Again, work by hand first, then check with:
Solution:
| > | int(2*Pi*(1-x)*exp(x),x=0..1); |
Arc Length and Surface Area
Problem: Calculate the length of the graph y=(1/2)*(exp(x)+exp(-x)) from x=0 to x=1.
Solution:
| > | restart: |
| > | f:=x->(1/2)*(exp(x)+exp(-x)); |
| > | D(f)(x); |
| > | expand(1+(D(f)(x))^2); |
| > | factor(simplify(1+(D(f)(x))^2)); |
| > | evalf(int(sqrt(%),x=0..1)); |
| > |
Problem: Calculate the length of the graph y=(1/3)*(x^2+2)^(3/2) from x=3 to x=6.
Solution:
| > | restart: |
| > | f:=x->(1/3)*(x^2+2)^(3/2); |
| > | D(f)(x); |
| > | expand(1+(D(f)(x))^2); |
| > | factor(simplify(1+(D(f)(x))^2)); |
| > | evalf(int(sqrt(%),x=3..6)); |
| > |
Problem: Calculate the area of the surface generated by revolving the graph of y=sqrt(x), from x=1 to x=2 about the x-axis. Then check your work:
Solution:
| > | restart: |
| > | f:=x->sqrt(x); |
| > | D(f)(x); |
| > | expand(1+(D(f)(x))^2); |
| > | factor(simplify(1+(D(f)(x))^2)); |
| > | simplify((2*Pi*sqrt(x))*sqrt((1+(D(f)(x))^2))); |
| > | evalf(int(%,x=1..2)); |
| > |
Problem: Calculate the lateral surface area of a cone of radius R and altitude H. Then check your work:
Solution:
| > | restart: |
First find an equation of the line passing through the origin and the point (R,H). We will revolve this line about the y-axis.
| > | f:=x->(H/R)*x; |
| > | D(f)(x); |
| > | expand(1+(D(f)(x))^2); |
| > | simplify(2*Pi*x*sqrt(1+(D(f)(x))^2)); |
| > | int(%,x=0..R); |
Notice that one R cancels out, and that the slant height of the cone is sqrt(R^2+L^2).
| > |
Work
Problem: A parabolic tank is formed by revolving the graph of y=x^2, x=0 to x=3, about the y-axis. The tank is initially full of oil weighing 50 pounds per cubic foot. Determine the amount of work required to pump all of the oil to a height of 10 feet above the top of the tank.
Solution:
| > | int(50*Pi*sqrt(y)^2*(19-y),y=0..9); |
| > |
Problem: A spherical tank has a radius of 5 feet. The tank contains oil weighing 45 pounds per cubic foot. The base of the tank is 20 feet above ground level. Determine the amount of work that was required to fill the tank from ground level.
Solution:
| > | int(45*Pi*(25-y^2)*(y+25),y=-5..5); |
| > |
Transcendental Functions
You have learned about the natural logarithm function and the exponential function f(x)=exp(x). Know their definitions, understand their properties, and be able to calculate derivatives and antiderivatives involving these functions.
Problem: Find f'(x) for the function f(x)=ln(e^x+e^(-x)).
| > | f:=x->ln(exp(x)+exp(-x)); |
| > | D(f)(x); |
Problem: Find f'(x) for the function f(x)=ln(sin(x)/x).
| > | f:=x->ln(sin(x)/x); |
| > | D(f)(x); |
Problem: Find dy/dx for the function y=x^(1+x^2). Note: do not forget about logarithmic differentiation. First take the logarithm of both sides of the equation, simplify the right side, differentiate both sides with respect to x, then solve for dy/dx:
| > | f:=x->x^(1+x^2); |
| > | D(f)(x); |
Problem: Find the volume generated by revolving the area under the graph of f(x)=exp(-x^2) about the y-axis.
Suggestion: Find the volume from x=0 to x=K, then let K approach infinity:
| > | eval(int(2*Pi*x*exp(-x^2),x=0..K)); |
| > | limit(%,K=infinity); |
| > |
Antidifferentiation Techniques
Evaluate the following antiderivatives by hand and check your answers:
1.
| > | int(sqrt(4-x^2),x)+C; |
2.
| > | int(x/sqrt(1+x^2),x)+C; |
3.
| > | int(x/(1+x^2),x)+C; |
4.
| > | int(x^2*sin(x),x)+C; |
5.
| > | int((5+sqrt(x))/(sqrt(x)),x)+C; |
6.
| > | int(x*ln(x),x)+C; |
7.
| > | int(cos(5*x)^2,x)+C; |
8.
| > | int(cos(5*x)^3,x)+C; |
9.
| > | int(x/sqrt(x+3),x)+C; |
10.
| > | int(exp(3*x)*cos(x),x)+C; |
11.
| > | int(sec(3*x)^4,x)+C; |
12.
| > | int((7*x^2+3*x+1)/(x^2-3*x+2),x)+C; |
13.
| > | int(1/(x^3-4*x^2+4*x),x)+C; |
14.
| > | int(sin(sqrt(x)),x)+C; |
15.
| > | int(ln(sqrt(x+3)),x)+C; |
16.
| > | int(3*x/(x^3-4*x^2),x)+C |
Of course, you can check other integrals in this way.
Differential Equations
Problem: You deposit $1000 into an account paying 8% annual interest, compounded continuously. In addition, you add to your investment by depositing money into the account continuously at the rate of $2000 per year. Set up and solve the differential equation whose solution represents the dollar value V(t) of your account after t years. The determine the dollar value of your account after 20 years.
Solution:
The differential equation is dV/dt = .08 * V + 2000, and the initial condition is V(0)=1000. Separate the variables and solve. Use the initial condition to determine the value of the constant of integration, and solve for V(t). You should obtain:
| > | restart: |
| > | V:=t->26000*exp(.08*t)-25000; |
So after 20 years, your money will be worth this many dollars:
| > | evalf(V(20)); |
| > |
Problem: You owe $5000 on your credit card, on which you pay 15% annual interest, compounded continuously. If you pay this debt continuously at the rate of $1000 per year, while not using your credit card, when will your credit card balance be zero?
Solution:
Let A represent the amount you owe after t years. The differential equation representing this situation is
dA/dt = .15*A-1000, with initial condition A(0)=5000. Separate the variables and solve. Be careful: .15*A-1000 is negative, so you must be sure to take the logarithm of the absolute value of .15*A-1000. Use the initial condition to determine the value of the constant of integration, and solve for A(t). You should obtain:
| > | restart: |
| > | A:=t->(1/.15)*(1000-250*exp(.15*t)); |
The debt will have been retired in this many years:
| > | fsolve(A(t)=0,t); |
How much will you pay during these years on this loan?
| > | 9.241962406*1000; |
How much do you owe at the end of one year?
| > | A(1); |
| > |