Basic Antiderivatives

Mean Value Theorem : (Page 289) If f is continuous on [a,b] and differentiable on (a,b) then there exists (at least one) a number c (a,b) such that f'( c ) = .

Proof: Depends on the extreme value theorem -- take Analysis I in university!

The next two results are very important in Integral Calculus (Cal II):

Theorem : If f'(x) = 0 for all x (a,b) then f is constant on (a,b).

Proof: Consequence of the MVT.

Corollary : If f'(x) = g'(x) for all x (a,b) then f(x) = g(x) + C where C is a constant.

Proof: Follows from previous result. (Do it!)

Most Important Definition of the Course : (pg. 351) A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I.

e.g. Show that and are both antiderivatives of on R.

Show that is an antiderivative of on R -{0} .

Note: F and G from the first e.g. above are two different antiderivatives of f but F(x) = G(x) + 5. As a result of the corollary above: any two antiderivatives of a function differ only by a constant. If F and G are two antiderivatives of f then G(x) = F(x) + C where C is a constant.

Theorem : (pg. 351) If F is any antiderivative of f on I then the most general antiderivative of f on I is F(x) + C where C is an arbitrary constant.

Note : Every differentiation gives us an antidifferentiation formula since if F'(x) = f(x) then F(x) + C is the (most general) antiderivative of f(x).

e.g. The Power Rule for Antiderivatives:

> f:=x^(n+1)/(n+1);Diff(f,x)=simplify(diff(f,x));

=> If n is any number then the antiderivative of is .

The following table should be memorized and proved . F and G stand for antiderivatives of f and g and c and n are . constants :

> restart:

e.g.

Find, and plot, three different antiderivatives of for x > 0.

> f:=3*x^2-x+1/x;F:=int(f,x);F1:=F+1;F2:=F-2;

> plot({f,F,F1,F2},x=0..3,y=-4..6,colour=black);

Notation: If F'(x) =f(x) then .

e.g. Find T. M. G. A. of the following functions:

> f1:=1+x+2*x^2-sqrt(x)+3*cos(x);f2:=(t^2-4*t^3)/t^(1/3);

> F1:=Int(f1,x)=int(f1,x)+C;F2:=Int(f2,t)=int(f2,t)+K;

e.g. Find f if f''(x) = and f(1) = 0 = f(2).

> fpp:=x^(-2);fp:=int(fpp,x)+C;f:=int(fp,x)+K;

> c1:=subs(x=1,f);c2:=subs(x=2,f);

> e1:=c1=0;e2:=c2=0;

> constants:=solve({e1,e2});

=>

#43, page 357

Given that the graph of f passes through (1,6) and that the slope of its tangent line at (x,f(x)) is 2x + 1, find f(2).

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

Find the value of C using the fact that (1,6) is on f.:

> C:=solve(subs(x=1,f)=6);

Find f(2) by plugging in!

> subs(x=2,f);

=> f(2) = 10.

> C:='C':

#65, page 358

A stone is dropped from the CN tower, 450 m above the gound. Answer the following asuming that the stone accelerates at -9.8 m/s.

a) Find the distance of the stone above ground level at time t.

b) How long does it take to reach the ground?

c) With what velocity does it strike the ground?

d) If the stown is thrown downward with a speed of 5 m/s, how long does it take to reach the ground?

> restart:a:=-9.8;v:=int(a,t)+C;d:=int(v,t)+K;

a) Find C and K using the facts that v(0) = 0 and d(0) = 450:

> C:=solve(subs(t=0,v)=0);K:=solve(subs(t=0,d)=450);

=> .

b) Solve d(t) = 0:

> sols:=solve(d=0,t);

=> ~9.6 seconds to hit ground.

c) plug in the appropriate value from (b) into v:

> subs(t=sols[2],v);

=> hits ground at about 94 m/s in the down direction (i.e. at -94 m/s).

d) Find the new v(t) using v(0) = -5 m/s.

> unassign('C');unassign('K');C:=solve(subs(t=0,v)=-5);K:=solve(subs(t=0,d)=450);d;

> solve(d=0,t);

=> It will hit this time after about 9.1 seconds.