The area problem and the definite integral

Area an definite Integral
Calculus I
Calculus II
Chemistry
Physics
Oceanography

By Rick Stoll

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 < ^...< x_n = b

with x_{i + 1} - x_i = $\Delta$ 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 = x_{i - 1}, and to the right by x = x_i, 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 [x_{i - 1}, x_i].

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

Area(L₁) + Area(L₂) + ^...+ Area(L_n) = $\displaystyle \Delta$ xf (x₀) + $\displaystyle \Delta$ xf (x₁) + ^...+ $\displaystyle \Delta$ xf (x_{n - 1})

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

Area(R₁) + Area(R₂) + ^...+ Area(R_n) = $\displaystyle \Delta$ xf (x₁) + $\displaystyle \Delta$ xf (x₂) + ^...+ $\displaystyle \Delta$ xf (x_n)

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 {x_i;i = 0, 1,^..., n} such that

x₀ = a < x₁ < x₂ < ^...< x_n = b

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

m_i = inf{f (x); x $\displaystyle \in$ [x_{i - 1}, x_i]} and M_i = sup{f (x); x $\displaystyle \in$ [x_{i - 1}, x_i]}

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

L_f(P) = m₁(x₁ - x₀) + m₂(x₂ - x₁) + ^...+ m_n(x_n - x_{n - 1})

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

U_f(P) = M₁(x₁ - x₀) + M₂(x₂ - x₁) + ^...+ M_n(x_n - x_{n - 1})

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

Theorem. We have

L_f(P) $\displaystyle \leq$ Area( $\displaystyle \Omega$ ) $\displaystyle \leq$ U_f(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

L_f(P) $\displaystyle \leq$ I $\displaystyle \leq$ U_f(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,^..., x_n} be a partition of [a, b]. Since f (x) = x² is increasing on [a, b], then m_i = x_{i - 1}² and M_i = x_i². So we have

U_f(P) = x²₁(x₁ - x₀) + x²₂(x₂ - x₁) + ^...+ x²_n(x_n - x_{n - 1})

and

L_f(P) = x²₀(x₁ - x₀) + x²₁(x₂ - x₁) + ^...+ x²_{n - 1}(x_n - x_{n - 1}) .

For each i, we have

x_{i - 1}² $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$ $\displaystyle \Big($ x_{i
- 1}² + x_{i - 1}x_i + x_i² $\displaystyle \Big)$ $\displaystyle \leq$ x_i²

since x_{i - 1} $\leq$ x_i. If we multiply by x_i - x_{i - 1}, we get

x_{i - 1}²(x_i - x_{i - 1}) $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$ $\displaystyle \Big($ x_{i
- 1}² + x_{i - 1}x_i + x_i² $\displaystyle \Big)$ (x_i - x_{i - 1}) $\displaystyle \leq$ x_i²(x_i - x_{i - 1}) .

But

$\displaystyle {\textstyle\frac{1}{3}}$ $\displaystyle \Big($ x_{i
- 1}² + x_{i - 1}x_i + x_i² $\displaystyle \Big)$ (x_i - x_{i - 1}) = x_i³ - x_{i
- 1}³

which implies

x_{i - 1}²(x_i - x_{i - 1}) $\displaystyle \leq$ $\displaystyle {\textstyle\frac{1}{3}}$ (x_i³ - x_{i - 1}³) $\displaystyle \leq$ x_i²(x_i - x_{i - 1}) .

Hence

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

since

(x₁³ - x₀³) + (x₂³ - x₁³) + ^...+ (x_n³ - x_{n - 1}³) = b³ - a³ .

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

$\displaystyle \int_{a}^{b}$ xⁿ dx = $\displaystyle {\frac{1}{n+1}}$ (b^{n
+ 1} - a^{n + 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 $\displaystyle \leq$ $\displaystyle {\textstyle\frac{9}{10}}$ ^.

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.

Send mail to m_engineer2002@netzero.com with questions or comments about this web site. Copyright © 2001 Math Depot Last modified: April 30, 2001

Send mail to m_engineer2002@netzero.com with questions or comments about this web site.
Copyright © 2001 Math Depot
Last modified: April 30, 2001