 Application Center - Maplesoft

# Fundamental Theorem of Calculus

You can switch back to the summary page by clicking here.

L26-fundamentalTheorem.mws

Calculus I

Lesson 26: The Fundamental Theorem of Calculus

We are going to continue the connection between the area problem and antidifferentiation. First we extend the area problem and the idea of using approximating rectangles for a continuous function which is not necessarily positive over the interval [a,b].

> restart; with(student):

> f:=1+5*x; > plot(f,x=-2..1); As we can see, the fucntion is not positive over the entire interval and if we "owned the land between the graph and the x axis" we would be thinking of the area of two triangles one below the x axis from -2 to -.2; the other above the axis from -.2 to 1. the first has area 1.8 *9*.5 and the other 1.2*6*.5 for a total of 11.7 square units. What happens when we use rightbox and sum for n=100 in this case?

> rightbox(f(x),x=-2..1,100); rightsum(f(x),x=-2..1,100); evalf(%);   We see that the answer is nowhere close to 11.7 and is in fact negative. Looking at the approximating rectangles shows us that in the sum when we take the height to be f(x*) for some x* in the subinterval , this value will be negative when x is between -2 and .2. So we guess that this process will be yielding the negative of the area of the first triangle + the area of the second triangle. That is 1.8*(-9)*.5 + 1.2*(4).5= -8.1 + 3.6=-4.5. This answer seems more consistent with the -4.275 above. This is considered a "Net Area" ie Area above x axis - Area belowbelow x axis. We switch to n=1000.

> rightbox(f(x),x=-2..1,1000); rightsum(f(x),x=-2..1,1000); evalf(%);   This seems to support the observation.

If g is continuous (or monotone) on [a,b] then this approximating rectangle sum process will have a limit which we denote as the definite integral of g over [a,b] the notation is and this definite integral can be interpreted as the difference of areas where g is above the x axis and below the x axis. We say that g is an integrable function when this happens.

We try this for our function > int(f,x=-2..1); This is the answer -4.5 that we were expecting. We also use the term ..."integrating f from -2 to 1" in describing this process of evaluating the definite integral. We can now define an indefinite integral which is a new function, based on this process of integrating continuous functions g, such as :

> G:=x-> Int( g(t),t=-2..x); This is defining a new function G for each x, which says to find its value you integrate g over the interval from -2 to x. The interesting fact is that this G is itself an antiderivative for g. That is G is differentiable and its derivative is g.

This is one part of the Fundamental theorem of Calculus. This yields a valuable tool in evaluating these definite integrals.

This says that over [a,b] G(b)-G(a) = This equation says that to find the definite integral, first we identify an antiderivative of g over [a, b] then simply evaluate that antiderivative at the two endpoints and subtract. Because of this theorem we can identify the process of finding antiderivatives and finding indefinite integrals. We check this with our we know an antiderivative would be .

> F:=int(f,x); You should note that here we have used the integral command int(f,x) without specifying the interval to get this antiderivative and we also gave it a name F.

> subs(x=-2,F); subs(x=1,F);subs(x=1,F)-subs(x=-2,F);   Here we see that we get the same answer as we expected -4.5. in our previous worksheet we found the area under the curve over the interval [0,3] to be 6. We check the process by integrating this expression over [0,3].

> int(x^2-4*x+5,x=0..3); We further check it by finding an antiderivative for the expression and applying the fundamental theorem.

> F:=int(x^2-4*x+5,x); subs(x=0,F); subs(x=3,F);subs(x=3,F)-subs(x=0,F);    This shows the antiderivative and its values at 0 and 3 and then substracts giving the same answer of 6.