Integration (Thomas-Finney),

Section 4.1, page 280 Problem 41

Evaluate .

Maple just does these.

>	Int(-3*(csc(x))^2,x)=int(-3*(csc(x))^2,x);

This is 3cot(x) of course, but note that Maple doesn't add the "+C" when it does anti-derivatives.

Section 4.3, page 297, Problem 52

Solve with initial condition .

There are two ways to solve this problem. First, we can just integrate (and put in the "+C" ourselves):

>	y:=int(4*x*(x^2+8)^(-1/3),x)+C;

Next, find C to satisfy the initial condition:

>	simplify(solve(subs(x=0,y)=0,C));

Now put this value of C back into y to get the answer:

>	subs(C=-12,y);

The other way to solve this problem is to use the "dsolve" command on the differential equation and initial condition as follows:

>

restart;

>	simplify(dsolve({diff(y(x),x)=4*x*(x^2+8)^(-1/3),y(0)=0},y(x)));

It's reassuriung that the answers are the same!

Section 4.5, page 320, Problem 20

Evaluate  .

Maple knows how to do sums, too:

>	Sum(k,k=1..13)=sum(k,k=1..13);

>	Sum(k^2,k=1..13)=sum(k^2,k=1..13);

>	Sum(k^3,k=1..13)=sum(k^3,k=1..13);

Problem 29

Graph the function over the interval  [0, 2]..  Partition the interval into four subintervals of equal length.and compute the Riemann sum in three ways: using the left end point of each subinterval, using the right end point and using the midpoint..

Maple has commands in the "student" library for drawing the rectangles associated

to various Riemann sums:

>

f:=x^2-1;

>	with(student):

>	leftbox(f,x=0..2,4);

>	middlebox(f,x=0..2,4);

>	rightbox(f,x=0..2,4);

We can also evaluate the corresponding Riemann sums:

>	leftsum(f,x=0..2,4),middlesum(f,x=0..2,4),rightsum(f,x=0..2,4);

and find their values:

>	value(leftsum(f,x=0..2,4)),value(middlesum(f,x=0..2,4)),value(rightsum(f,x=0..2,4));

Compare these to the actual integral:

>	int(f,x=0..2);

We'll need more rectangles to get a better approximation.

Section 4.6, page 330, Problem 30

Graph on [0,1] and compute its average value.  At what point(s) in the interval does the function assume its average value?

The average value of a function on an interval is its integral divided by the length of the interval. So for this problem,

>	avg:=int(3*x^2-3,x=0..1)/(1-0);

We are also asked to find where the function assumes its average:

>	solve(3*x^2-3=avg,x);

Only the first of these is in the interval, however.

Section 4.7, page 342, Problem 85

Let for x in the interval and let .

a) Plot both f and F over .

b) Solve the equation  F'(x)=0.  What does the Fundamental Theorem of Calculus say about this?

c) Over what intervals is the function (approximately) increasing or decreasing?  What is true about f in these intervals?

(a) We start by defining the requisite functions:

>	f:=x->sin(2*x)*cos(x/3);

>	F:=x->int(f(t),t=0..x);

>	plot({f(x),F(x)},x=0..2*Pi,color=blue,thickness=2);

>

G:=D(F);

Note that Maple knows the fundamental theorem! Since F was defined as the integral of f, it gave immediately that the derivative of F is f.

>

G(x);

(b)

>	solve(G(x)=0);

Maple gives only one solution of G(x)=0, but we suspect there are more (we saw them on the graph -- one between 1 and 2, another between 3 and 4, perhaps another between 4 and 5), since G(x) is a periodic, trigonometric function. Let's try "fsolving" for them:

>	fsolve(G(x)=0,x=1..2);fsolve(G(x)=0,x=3..4);fsolve(G(x)=0,x=4..5);

That looks like , and .

(c) Of course, since f is the derivative of F, we see that f is positive when F is increasing (check out the graph). So F is increasing on , and then on .

(d) The derivative of f is the second derivative of F, so f ' should reflect the concavity of F, and zeroes of f ' should be inflection points:

>	plot({diff(f(x),x),F(x)},x=0..2*Pi,color=blue,thickness=2);

Section 4.8, page 345, Problem 27

Find the area between the curve  and the x axis for x in the interval .

To find the area, we need to integrate the function -y over the region, since y is negative:

>	y:=3*sin(x)*sqrt(1+cos(x));

>	plot(y,x=-Pi..0,color=blue,thickness=2);

>	area:=int(-y,x=-Pi..0);

Section 4.9, page 353, Problem 7

Estimate   with a subdivision into 4 intervals using a) the trapezoidal rule and b) Simpson's rule.  In both cases, find an estimate for the Error.

We will use the commands "trapezoid" and "simpson" from the student library:

>	with(student,trapezoid,simpson):

Let's calculate the actual integral first:

>	exact:=int(1/s^2,s=1..2);

OK, next up is the trapezoidal rule:

>	tapp:=trapezoid(1/s^2,s=1..2,4);

>

value(%);

>	tapp:=evalf(%);

Pretty close, how about Simpson's rule?

>	sapp:=evalf(simpson(1/s^2,s=1..2,4));

Even better.  Now for error estimates. For trapezoidal error we need an upper bound on the second derivative of the integrand, and for Simpson's error we need a bound on the fourth derivative:

>	diff(1/s^2,s$2); diff(1/s^2,s$4);

Since these are decreasing functions for positive s, the upper bounds occur at the left point of our interval, i.e., at s=1. Now we plug into formulas (3) and (7) for the error bouds:

>	trapbound:=(2-1)/12*((2-1)/4)^2*6; simpbound:=(2-1)/180*((2-1)/4)^4*120;

>	evalf(trapbound),evalf(simpbound);

We can see that these bounds are bigger than the actual errors, which are:

>	tapp-exact,sapp-exact;

The actual percentage error in the calculations are:

>	trappcterr:=(tapp-exact)/exact*100; simppcterr:=(sapp-exact)/exact*100;