Introduction: The Area Problem and the Definite Integral


Integration is vital to many scientific areas. Many powerful mathematical tools are based on integration. Differential equations for instance are the direct consequence of the development of integration.

So what is integration? Integration stems from two different problems. The more immediate problem is to find the inverse transform of the derivative. This concept is known as finding the antiderivative. The other problem deals with areas and how to find them. The bridge between these two different problems is the Fundamental Theorem of Calculus.

What is the "area problem"? We want to find the area of a given region in the plane. It is not hard to see that this problem can be reduced to finding the area of the region $ \Omega$ bounded above by the graph of a positive function f (x), bounded below by the x-axis, bounded to the left by the vertical line x = a, and to the right by the vertical line x = b.

The answer to this problem came through a very nice idea. Indeed, let us split the region $ \Omega$ into small subregions which we can approximate by rectangles or other simple geometrical figures (whose areas we know how to compute). This is how it goes: split the interval [a, b] into subintervals, preferably with the same width $ \Delta$x,

x0 = a < x1 < x2 < ... < xn = b

with xi + 1 - xi = $ \Delta$x = $\displaystyle {\frac{b-a}{n}}$, for i = 0, 1, ... , n - 1.

Let $ \Omega_{i}^{}$ be the subregion bounded above by the graph of f (x), bounded below by the x-axis, bounded to the left by x = xi - 1, and to the right by x = xi, for i = 1, ... , n. Clearly we have

Area($\displaystyle \Omega$) = Area($\displaystyle \Omega_{1}^{}$) + Area($\displaystyle \Omega_{2}^{}$) + ... + Area($\displaystyle \Omega_{n}^{}$)  .

So we focus on the subregions $ \Omega_{i}^{}$, for i = 1, ... , n. Since we want to approximate the regions by rectangles, then we only have to worry about the upper boundary of each region (since on the other sides we already have straight lines). Again: We are looking for good approximations of the regions $ \Omega_{i}^{}$ by rectangles.

The easiest way to choose a height for our rectangles is to choose the value of the function at the left (or right) end points of the small intervals [xi - 1, xi].

Let Li be the rectangle defined by the left-end point and Ri be the rectangle defined by the right-end point. Then an approximation to Area($ \Omega$) is given by

Area(L1) + Area(L2) + ... + Area(Ln) = $\displaystyle \Delta$xf (x0) + $\displaystyle \Delta$xf (x1) + ... + $\displaystyle \Delta$xf (xn - 1)

which we will call the left-sum denoted LEFT(n), and

Area(R1) + Area(R2) + ... + Area(Rn) = $\displaystyle \Delta$xf (x1) + $\displaystyle \Delta$xf (x2) + ... + $\displaystyle \Delta$xf (xn)

which we will call the right-sum denoted RIGHT(n)

Example. Consider the function

f (x) = x2

for x $ \in$ [0, 1]. Let us split the interval into 4 subintervals. We have

x0 = 0  ,  x1 = $\displaystyle {\textstyle\frac{1}{4}}$  ,  x2 = $\displaystyle {\textstyle\frac{1}{2}}$  ,  x3 = $\displaystyle {\textstyle\frac{3}{4}}$  ,  x4 = 1  .

We have $ \Delta$x = $\displaystyle {\textstyle\frac{1}{4}}$ and

LEFT(4) = $\displaystyle {\textstyle\frac{1}{4}}$02 + $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{1}{4}}\right.$$\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left.\vphantom{\frac{1}{4}}\right)^{2}_{}$ + $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{1}{2}}\right.$$\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \left.\vphantom{\frac{1}{2}}\right)^{2}_{}$ + $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{3}{4}}\right.$$\displaystyle {\textstyle\frac{3}{4}}$$\displaystyle \left.\vphantom{\frac{3}{4}}\right)^{2}_{}$ = $\displaystyle {\textstyle\frac{7}{32}}$

and

RIGHT(4) = $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{1}{4}}\right.$$\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left.\vphantom{\frac{1}{4}}\right)^{2}_{}$ + $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{1}{2}}\right.$$\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \left.\vphantom{\frac{1}{2}}\right)^{2}_{}$ + $\displaystyle {\textstyle\frac{1}{4}}$$\displaystyle \left(\vphantom{\frac{3}{4}}\right.$$\displaystyle {\textstyle\frac{3}{4}}$$\displaystyle \left.\vphantom{\frac{3}{4}}\right)^{2}_{}$ + $\displaystyle {\textstyle\frac{1}{4}}$12 = $\displaystyle {\textstyle\frac{15}{32}}$  .

Note that Area($ \Omega$) equals $\displaystyle {\textstyle\frac{1}{3}}$, a result which we will prove in later pages.

Indeed if the function f (x) is not too badly behaved, we will show that when n gets larger, the numbers LEFT(n) and RIGHT(n) get closer to Area($ \Omega$), i.e.

Area($\displaystyle \Omega$) = $\displaystyle \lim_{n \rightarrow \infty}^{}$LEFT(n) = $\displaystyle \lim_{n \rightarrow \infty}^{}$RIGHT(n)  .

This is the main idea described above. The number Area($ \Omega$) is called the definite integral (or more simply the integral) of f (x) from a to b and is denoted by

$\displaystyle \int_{a}^{b}$f (x) dx  .

Note that in the expression $ \int_{a}^{b}$f (x) dx the variable x may be replaced by any other variable.

Example. Let $ \alpha$ $ \geq$ 0. Then we have

$\displaystyle \int_{a}^{b}$$\displaystyle \alpha$ dx = $\displaystyle \alpha$(b - a)  .

This is true since the region $ \Omega$ is simply a rectangle.

Example. We have

$\displaystyle \int_{a}^{b}$x dx = $\displaystyle {\textstyle\frac{1}{2}}$(b2 - a2)  .

Indeed, the region $ \Omega$ is simply the union of two regions: one rectangle and one triangle.

The rectangle (depicted in red) is bounded above by x = a and its area is a(b - a). The triangle (in blue) is determined by the points: (a, a), (a, b), and (b, b). Its area is $\displaystyle {\textstyle\frac{1}{2}}$(b - a)2. So we have

$\displaystyle \int_{a}^{b}$xdx = a(b - a) + $\displaystyle {\textstyle\frac{1}{2}}$(b - a)2 = $\displaystyle {\textstyle\frac{1}{2}}$(b2 - a2)  .

A precise definition for the definite integral involves partitions and lower as well as upper sums:

Definition. A partition P of the interval [a, b] is a sequence of numbers {xi;i = 0, 1, ... , n} such that

x0 = a < x1 < x2 < ... < xn = b

For a function f (x) defined on [a, b] and a partition P of [a, b], set

mi = inf{f (x);  x $\displaystyle \in$ [xi - 1, xi]}    and    Mi = sup{f (x);  x $\displaystyle \in$ [xi - 1, xi]}

for i = 1, ... , n, provided that f (x) is bounded on [a, b]. The sum

Lf(P) = m1(x1 - x0) + m2(x2 - x1) + ... + mn(xn - xn - 1)

is called the lower sum for f (x) over the partition P, and

Uf(P) = M1(x1 - x0) + M2(x2 - x1) + ... + Mn(xn - xn - 1)

is called the upper sum for f (x) over the partition P.

Theorem. We have

Lf(P) $\displaystyle \leq$ Area($\displaystyle \Omega$) $\displaystyle \leq$ Uf(P)

for any partition P of [a, b]. Moreover if f (x) is continuous on [a, b], except maybe at a finite number of points, and I is a number such that

Lf(P) $\displaystyle \leq$ I $\displaystyle \leq$ Uf(P)

for any partition P of [a, b], then I = Area($ \Omega$).

This theorem is fundamental. Let us illustrate this with the following example.

Example. Use the above theorem to show

$\displaystyle \int_{a}^{b}$x2dx = $\displaystyle {\textstyle\frac{1}{3}}$(b3 - a3)  

where b $ \geq$ a $ \geq$ 0. Let P = {x0, x1, ... , xn} be a partition of [a, b]. Since f (x) = x2 is increasing on [a, b], then mi = xi - 12 and Mi = xi2. So we have

Uf(P) = x21(x1 - x0) + x22(x2 - x1) + ... + x2n(xn - xn - 1)

and

Lf(P) = x20(x1 - x0) + x21(x2 - x1) + ... + x2n - 1(xn - xn - 1)  .

For each i, we have

xi - 12 $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$$\displaystyle \Big($xi - 12 + xi - 1xi + xi2$\displaystyle \Big)$ $\displaystyle \leq$ xi2

since xi - 1 $ \leq$ xi. If we multiply by xi - xi - 1, we get

xi - 12(xi - xi - 1) $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$$\displaystyle \Big($xi - 12 + xi - 1xi + xi2$\displaystyle \Big)$(xi - xi - 1) $\displaystyle \leq$ xi2(xi - xi - 1)  .

But

$\displaystyle {\textstyle\frac{1}{3}}$$\displaystyle \Big($xi - 12 + xi - 1xi + xi2$\displaystyle \Big)$(xi - xi - 1) = xi3 - xi - 13

which implies

xi - 12(xi - xi - 1) $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$(xi3 - xi - 13) $\displaystyle \leq$ xi2(xi - xi - 1)  .

Hence

Lf(P) $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$(b3 - a3) $\displaystyle \leq$ Uf(P)

since

(x13 - x03) + (x23 - x13) + ... + (xn3 - xn - 13) = b3 - a3  .

Exercise 1. Use similar ideas as used in the example above to show

$\displaystyle \int_{a}^{b}$xn dx = $\displaystyle {\frac{1}{n+1}}$(bn + 1 - an + 1)  

where b $ \geq$ a $ \geq$ 0.

Exercise 2. Show that

$\displaystyle {\textstyle\frac{13}{20}}$ $\displaystyle \leq$ $\displaystyle \int_{0}^{1}$$\displaystyle {\frac{1}{1+x^2}}$ dx $\displaystyle \leq$ $\displaystyle {\textstyle\frac{9}{10}}$   .

Answer.

Exercise 3. Consider the function

f (x) = $\displaystyle \left\{\vphantom{\begin{array}{lll}
1 & \mbox{if $x$ is rational}\\
0 & \mbox{if $x$ is irrational .}\\
\end{array}}\right.$$\displaystyle \begin{array}{lll}
1 & \mbox{if $x$ is rational}\\
0 & \mbox{if $x$ is irrational .}\\
\end{array}$

Show that $ \int_{0}^{1}$f (x) dx does not exist.

Answer.


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