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We develop the theory of integration of continuous functions and prove the fundamental theorem of calculus.

Let R be any (bounded) region in the plane which we want to measure the area of. We can tile the plane with squares of unit length and count the number of such squares that are contained in the region to obtain an approximation to the area from below. On the other hand we can count the number of squares that meet to region to obtain an approximation to the area from above. We can repeat this with squares of smaller size and appropriately scale the count it seems clear that the approximant from below will increase and the approximant from above will decrease. The least upper bound of the former is called the inner measure and the greatest lower bound of the latter the outer measure. To obtain an area for the region we must show that these two numbers are the same; moreover, we would like these numbers to be independent of the placement of the grid as well as rotation and/or shearing of the grid.

Exercise 53   Show that the measure of any rectangle is the product of the two sides. More generally, show that the area of a parallelogram is the product of the height and the base.

Let f (x) be a (non-negative) continuous function for a $ \leq$ x $ \leq$ b. Let R be the region bounded by the lines x = a and x = b on the left and right, by the x-axis below and the curve y = f (x) at the top. By the previous exercise we can calculate the measure by using rectangles instead of squares. We do this in the following exercise.

Exercise 54   By a partition P of the interval [a, b] we mean a (finite) collection of points a = t0 < t1 < ... < tn = b. For any such partition we define mi (respectively Mi) to be the minimum (respectively maximum) value of f (x) for ti - 1 $ \leq$ x $ \leq$ ti.
  1. Show that the sum L(P, f ) (respectively U(P, f )) approximate the area of the region from below (respectively above), i. e. are sums of areas of rectangles enclosed by (respectively enclosing) the region R.
    L(P, f ) = $\displaystyle \sum_{i=1}^{n}$mi(ti - ti - 1)  
    U(P, f ) = $\displaystyle \sum_{i=1}^{n}$Mi(ti - ti - 1)  

  2. If P' is a finer partition than P (i. e. each point of P is also a point of P') then show that

    L(P, f ) $\displaystyle \leq$ L(P', f ) $\displaystyle \leq$ U(P', f ) $\displaystyle \leq$ U(P, f )

  3. Let Pn denote the partition of [a, b] into n equal parts. Show that

    sup{L(P, f )| P a partition } $\displaystyle \geq$ sup{L(Pn, f )}

    Similarly for the infimum of the U(P, f ),

    inf{U(P, f )| P a partition } $\displaystyle \leq$ inf{U(Pn, f )}

  4. Let i = i(P, f ) be such that the difference Mi - mi is maximum. Then show that

    U(P, f )- L(P, f ) $\displaystyle \leq$ (Mi - mi)(b - a)

    Let x(P, f ) denote the mid point of the interval [ii - 1, ti] for this i.
  5. Let c be any point of the interval [a, b]. For any positive $ \epsilon$, show that there is a $ \delta$ > 0 so that the difference between the maximum and minimum values of f (x) on the interval [c - $ \delta$, c + $ \delta$] is less than $ \epsilon$/(b - a). (Hint: use continuity of f at c).
  6. The sequence {xn = x(Pn, f )} has a convergent subsequence {yk = xnk}, with limit point c. Show that there is a k0 so that if k $ \geq$ k0 and i = i(Pnk, f ) then the entire sub-interval [ti - 1, ti] of the partition Pnk is contained in [c - $ \delta$, c + $ \delta$].
  7. Deduce that sup{L(Pnk, f )} = inf{U(Pnk, f )}.
  8. Conclude that the inner and outer measure of the region R coincide.

This show that the area of the region R is well-defined. It is denoted by $ \int_{a}^{b}$f to denote its depends on a, b and f (we will justify the use of the $ \int$ symbol below. First of all note that

Exercise 55   Let a < c < b and f (x) and g(x) be (positive) functions that are continuous in the interval [a, b]; let d > 0 be any positive constant. Then we have

$\displaystyle \int_{a}^{c}$f + $\displaystyle \int_{c}^{b}$f = $\displaystyle \int_{a}^{b}$f


$\displaystyle \int_{a}^{b}$(d . f + g) = d . $\displaystyle \int_{a}^{b}$f + $\displaystyle \int_{a}^{b}$g

Due to this we are justified in extending the definition as follows. If f (x) is continuous in the interval [a, b] and c > d are points in this interval we define $ \int_{c}^{d}$f : = - $ \int_{d}^{c}$f. Moreover, if f is not everywhere positive we define

f+(x) = max{f (x,), 0} = (f (x) + | f (x)|)/2


f-(x) = max{ - f (x,), 0} = (- f (x) + | f (x)|)/2

Then clearly f± are positive functions and f = f+ - f-. We then define the integral of f by the formula $ \int_{a}^{b}$f : = $ \int_{a}^{b}$f+ - $ \int_{a}^{b}$f-. This definition too is justified by the additive property given above.

Now if f is continuous on an interval [a, b] where its minimum value is m and its maximum value is M it is clear that

m(d - c) $\displaystyle \leq$ $\displaystyle \int_{c}^{d}$f $\displaystyle \leq$ M(d - c)

Exercise 56   Let g(x) = $ \int_{a}^{x}$f, then g is a differentiable function for all points x so that a < x < b and its derivative is f. (Hint: Apply the addition rule and use continuity of f at x)

This justifies the use of the $ \int$ symbol. We have shown how to analytically compute the function which we formally defined above. In addition we have a way of constructing a function which is differentiable and its derivative is continuous--a problem that was raised in the section on continuous and differentiable functions.

next up previous
Next: Curves Up: New Functions from old Previous: Implicit functions
Kapil H. Paranjape 2001-01-20