Area

Area is a concept familiar to all of us from our study of various geometric figures such as rectangles, triangles, and circles. We generally think of area as a number that in some way suggests the size of a bounded region. Of course, we have specific formulas for calculating the areas of simple geometric figures.
Look also at the Fairy Tale - Thoughts for a Wedding Reception.

Our problem here is to develop a way to calculate the area of a plane region R, bounded by the x-axis, the lines x = a and x = b, and the graph of a nonnegative continuous function f, as shown in the next Figure.


The solution to this problem is given by the Fundamental Theorem of Calculus and represents one of the most famous discoveries in mathematics. From this theorem we can see that, just as the derivative can be used to find slope, the antiderivative can be used to find area. In anticipation of the connection between antiderivatives and area, we denote the area of the region shown in the previous Figure by

∫ a b f ( x )   d x

The symbol ∫ a b   f ( x )   d x   is called the definite integral from a to b, where a is the lower limit of integration and b is the upper limit of integration.


Definite Integrals and the Fundamental Theorem of Calculus

In developing the Fundamental Theorem of Calculus, we temporarily introduce the area function shown in the next Figure


Specifically, if f is continuous and nonnegative on [a,b], we denote the area of the region under graph of f from a to x by A(x).

Now, if we let x increase by an amount △ x , then the area of the region under the graph of f increases by △ A . Furthermore, if f(m) and f(M) denote the minimum and the maximum values of f on the interval [ x , x + △ x ] , then we have the following relationship

f( m )   △ x ⏞ Area     of inscribed rectangle     ≤     △ A     ≤ f( M )   △x ⏞ Area     of circumscribed rectangle


and


Dividing each term in the inequality
f ( m ) △ x ≤   △ A ≤ f ( M ) △ x
by the positive number △ x , we have
f ( m ) ≤ △ A △ x ≤ f ( M )

Since both  f(m) and  f(M) approach  f(x) as △ x   approaches zero and since
lim △ x &to; 0 △ A △ x = A ′ ( x )
it follows that
f ( x ) ≤ A ′ ( x ) ≤ f ( x )
Which means that
f(x) = A' (x)
Thus, we have established that the area function  A(x) is an antiderivative of f and consequently must be of the form
A (x) = F(x) + C
where  F(x) is any antiderivative of  f. To solve for  C, we note
that  A(a) = 0 and thus  C = -F(a) (see also Particular Solution web page). Furthermore, evaluating  A(b), we have
A(b) = F(b) + C = F(b) -F(a)

Finally, replacing  A(b) by its integral form, we have
∫ a b f ( x ) = F ( b ) - F ( a )

This equation tells us that if we can find an antiderivative for  f, then we can use the antiderivative to evaluate the definite integral ∫ a b f ( x ) d x . We summarize this result in the following theorem.

Fundamental Theorem of Calculus
If a function  f is nonnegative and continuous on the closed interval  [a,b] then
∫ a b f ( x ) = F ( b ) - F ( a )
where  F is any function such that  F ' (x) = f (x) for all  x in  [a,b].