STUDENT >
restart;
STUDENT >
with(plots):
STUDENT >
with(student):
Recall
that Maple allows us to do many operations quickly and effeciently. One of
these is integration. If we identify a function we can see its graph, and then
integrate it over any given interval.
STUDENT >
plot(x^3+1,x=-2..2,color=black);
STUDENT >
Int(x^3+1,x=-1..2);
STUDENT >
value(");
Note
how Maple treats the following problem.
STUDENT >
Int(x^3+1,x);
STUDENT >
value(");
Maple
does not put in the constant of integration. It also has problems integration
some expressions such as the following function, even though it can give us a
good approximation of the definite integral.
STUDENT >
Int(sqrt(x^7+4),x);
STUDENT >
Int(sqrt(x^7+4),x=1..3);
STUDENT >
evalf(");
We will
start by finding the area between a curve and the x axis. To help us do this,
we will graph the function.
STUDENT >
f:=5*x^3-19*x^2+11*x+3;
STUDENT >
plot(f,x=-2..5,color=black);
Since
the curve crosses the x-axis several times we must find those intersection
points. Then using our rules for integration, we can set up the appropriate
integrals.
STUDENT >
pt:=fsolve(f=0,x);
Since
the graph is above the x-axis from [-1/5,1] and falls below the x-axis on the
interval [1,3] we must calculate the total area by adding the area calculated
by the first integral to the negative of that calculated by the second
integral.
STUDENT >
Int(f,x=pt[1]..pt[2])-Int(f,x=pt[2]..pt[3]);
STUDENT >
value(");
We can
use the same technique to find the area between two curves. Remember to graph
the curves so that you always know which equation is the top curve ( or the
rightmost curve).
STUDENT >
f:=3*x^3-x^2-10*x; g:=-x^2+2*x;
STUDENT >
plot({f,g},x=-4..4);
STUDENT >
pt:=fsolve(f=g,x);
STUDENT >
Int(f-g,x=pt[1]..pt[2])+Int(g-f,x=pt[2]..pt[3]);
STUDENT >
value(");
1. Find
the area between the given curve and the x-axis. Use the graph to help you.
4 3 2
y := x + .63 x - 15.456 x + 12.853 x + 5
2.
Graph the region bounded by the given curves, and use the graph and Maple to
help calculate the area of the region.
3 1/2
y[1] := (1 + x )
y[2] := 1/2 x + 2
x := 0
3. Find
the area of the region that is bounded above by the following curves, and below
by the x-axis.
3
y[1] := 4 - x - x
y[2] := 7 ln(x)
STUDENT >
restart; with(plots): with(student):
Just
like area we must get an idea how to approach a volume problem, so that we can
set it up for Maple to solve. Let's first solve a problem using the washer
method. To do this we will revolve the following equations about the x-axis.
STUDENT >
f:=sqrt(x);g:=x^2;
STUDENT >
plot({f,g},x=-1..2);
STUDENT >
pt:=fsolve(f=g,x);
Notice
that Maple failed to find the point of intersection (0,0). This is a common
problem when the intersection point is the endpoint on one curve. Always check
the answers Maple gives you!
STUDENT >
Int(Pi*(f^2-g^2),x=0..pt);
STUDENT >
value(");
Let's
take the same problem and integrate using the shell method. Thanks to Maple, we
have a graph that allows us to set up the problem easily. We even know the
intersection points for these two curves and the position of each. However, our
integration will now be with respect to y.
STUDENT >
Int(2*Pi*y*(sqrt(y)-y^2),y=0..1);
STUDENT >
value(");
Notice
when we use Maple's calculated answers it gives us an approximated answer,
however with exact limits of integration Maple gives an exact answer. Use
evaluate to make sure our results are the same.
STUDENT >
evalf(");
Use
each of the three methods used to find volume (one per problem) to calculate
the volume of the solid generated by revolving the region bounded by the
curves:
3
y[1] := x
y[2] := 0
x := 2
1.
about the x-axis
2.
about the y-axis
3.
about the line x=4