The Fundamental Theorem of Calculus (F.T.C.)

Part 1.

The FTC part 1 : If f is continuous on [a,b] then for x [a,b] is continuous on [a,b] and differentiable on (a,b) and g'(x) = f(x) .

Proof: Hard. See board.

e.g. Find g'(x) two ways if g(x) = .

> g:=Int(sqrt(1+t^2),t=0..x);

> diff(g,x);

> gg:=value(g);

> diff(gg,x);

> simplify(%);

e.g. Find h'(x) if h(x) =

> h:=Int(sin(p^3),p = x .. -2);diff(h,x);

e.g. Find F'(x) if F(x) =

> F:=Int(ln(t^2+1),t = 0 .. x^3);

> diff(F,x);

e.g. Find G'(x) if G(x) =

> G:=Int(ln(t^2+1),t = x^2 .. x^3);

> diff(G,x);

Part 2.

The FTC part 2 : If f is continuous on [a,b] and F is any antiderivative of f on [a,b] then

Proof: Follows quickly from Part 1 .

e.g.'s Use the FTC to confirm the area calculations from the Previous Section

1. Find the area under y = 2x + 1 and above [0,4]

=> Find

> f:=2*x+1;F:=int(f,x);

> Area=subs(x=4,F)-subs(x=0,F);

> Int(2*x+1,x=0..4)=int(2*x+1,x=0..4);

2. Find

> Int(x^2+2,x=1..2)=int(x^2+2,x=1..2);

3. Find

> Int(4*x-x^3,x = 0 .. 3)=int(4*x-x^3,x = 0 .. 3);

Evaluate the following definite integrals :

> f1:=Int(3/(t^4),t=1..2);f2:=Int(sqrt(x^2-5),x=4..4);f3:=Int(sin(q),q=Pi/4..Pi/3);f4:=Int(6/(1+x^2),x=1..sqrt(3));

> value(f1);value(f2);value(f3);value(f4);

> f5:=piecewise(0<=x and x<1,x^4,1<=x and x<=2,x^5);plot(f5,x=0..3,y=0..30,colour=black);

> Int(f5,x=0..2)=int(f5,x=0..2);

Total Change

By the FTC, if F'(x) = f(x) (i.e. if F is an antiderivative of f) then where f is the rate of change of F. It follows that we can interprete the definite integral as the (total) change in F with respect to x as x changes from a to b.

For example, if v(t) is the rate of change of position (i.e. the velocity) of an object moving in a straight line then is its total change in position (i.e. its displacement). Note: this is not the same as the total distance travelled by the object. Look at a sketch of a velocity curve to see why the total distance travelled would be .

This idea of total change can be applied to any functional relationship involving the FTC: .

e.g.

(#55, page 408) Suppose a particle moving in a straight line is accelerating (in m/ ) according to , , t [0,10] (this would not happen in Nature!). Find (a) the velocity at time t and (b) the distance travelled.

> a:=t+4;v:=int(a,t)+C;

Now, v(0) = 5 so C = 5:

> C:=5;v;

=>

> plot(v,t,colour=black);t_intercepts:=solve(v=0);

You can see from the plot (and by knowing that the graph is parabola with a minimum point and the t-intercepts as above) that the velocity is positive for so the total displacement over [0,10] is the same as the total distance travelled over the same interval. Therefore:

> total_distance:=Int(v,t=0..10);

> value(total_distance);

> evalf(%);

=> The particle covers a distance of ~417 meters.

e.g. A particle moving in si mple harminic motion.

Suppose a mass attached to a spring is oscillating at the rate units/s. Determine its displacement after 1 second and the distance it travels in the first second.

> with(plots):

> v:=t->-8/5*sin(8*t);

> plot1:=plot(v(t),t=0..1,colour=blue,style=point):plot2:=plot(abs(v(t)),t=0..1,thickness=2):

> display({plot1,plot2},title=`velocity(broken) and speed (line)`);

> int(v(t),t);

> displ:=int(v(t),t=0..1);evalf(%);

=> Its displacement at t = 1 is ~0.23 units (i.e. at 1 second it is 0.29 units to the left of equilibrium.

> dist:=int(abs(v(t)),t=0..1);evalf(%);

=> It travels a total of ~1.03 units in 1 second.

Alternately, if you don't know the antiderivative of you would proceed as follows:

> solve(abs(v(t))=0);

Maple is not much help in finding the intercepts of v, so you solve the equation "by hand" to get the first two nonzero intercepts:

> intercepts:=seq(k*Pi/8,k=1..2);

> Int(-v(t),t=0..intercepts[1])+Int(v(t),t=intercepts[1]..intercepts[2])+Int(-v(t),t=intercepts[2]..1)=value(%);;

> evalf(rhs(%));

=> Same answer! (phew!)