AOS3 Topic 11: Application of Integration
In this section we look at three applications of integration.
Average value of a function
The average value of a function \(f\) for an interval \([a, b]\) is defined as:
\(\frac{1}{b - a} \int_a^b f(x) dx\)
In terms of the graph of \(y = f(x)\), the average value is the height of a rectangle having the same area as the area under the graph for the interval \([a, b]\).
Rates of Change
Given the rate of change of a quantity we can obtain information about how the quantity varies. For example, we have seen that if the velocity of an object travelling in a straight line is given at time t, then the position of the object at time t can be determined using information about the initial position of the object.
The Fundamental Theorem of Calculus
The derivative of the area function
Let \(f : [a, b] \to \mathbb{R}\) be a continuous function such that \(f(x) \geq 0\) for all \(x \in [a, b]\).
We define the function \(A\) geometrically by saying that \(A(x)\) is the measure of the area under the curve \(y = f(x)\) between \(a\) and \(x\). We thus have \(A(a) = 0\). We will see that \(A'(x) = f(x)\), and thus \(A\) is an antiderivative of \(f\).
First consider the quotient \(\frac{A(x + h) - A(x)}{h}\) for \(h > 0\).
By our definition of \(A(x)\), it follows that \(A(x + h) - A(x)\) is the area between \(x\) and \(x + h\).
Let \(c\) be the point in the interval \([x, x+h]\) such that \(f(c) \geq f(z)\) for all \(z \in [x, x + h]\), and let \(d\) be the point in the same interval such that \(f(d) \leq f(z)\) for all \(z \in [x, x + h]\).
Thus \(f(d) \leq f(z) \leq f(c)\) for all \(z \in [x, x + h]\).
Therefore \(hf(d) \leq A(x + h) - A(x) \leq hf(c)\).
That is, the shaded region has an area less than the area of the rectangle with base \(h\) and height \(f(c)\) and an area greater than the area of the rectangle with base \(h\) and height \(f(d)\).
Dividing by \(h\) gives
\(f(d) \leq \frac{A(x + h) - A(x)}{h} \leq f(c)\)
As \(h \to 0\), both \(f(c)\) and \(f(d)\) approach \(f(x)\).
Thus we have shown that \(A'(x) = f(x)\), and therefore \(A\) is an antiderivative of \(f\).
Now let \(G\) be any antiderivative of \(f\). Since both \(A\) and \(G\) are antiderivatives of \(f\), they must differ by a constant. That is,
\(A(x) = G(x) + k\)
where \(k\) is a constant. First let \(x = a\). We then have
\(0 = A(a) = G(a) + k\)
and so \(k = -G(a)\).
Thus \(A(x) = G(x) - G(a)\), and letting \(x = b\) yields
\(A(b) = G(b) - G(a)\)
The area under the curve \(y = f(x)\) between \(x = a\) and \(x = b\) is equal to \(G(b) - G(a)\), where \(G\) is any antiderivative of \(f\).
A similar argument could be used if \(f(x) \leq 0\) for all \(x \in [a, b]\), but in this case we must take \(A(x)\) to be the negative of the area under the curve. In general:
Fundamental theorem of calculus
If \(f\) is a continuous function on an interval \([a, b]\), then
\(\int_a^b f(x) dx = G(b) - G(a)\)
where \(G\) is any antiderivative of \(f\).
The area as the limit of a sum
Finally, we consider the limit of a sum in a special case. This discussion gives an indication of how the limiting process can be undertaken in general.
Notation
We first introduce a notation to help us express sums. We do this through examples:
\(\sum_{i=1}^3 i^2 = 1^2 + 2^2 + 3^2\)
\(\sum_{i=1}^5 x_i = x_1 + x_2 + x_3 + x_4 + x_5\)
\(\sum_{i=1}^n x_i f(x_i) = x_1 f(x_1) + x_2 f(x_2) + x_3 f(x_3) + \cdots + x_n f(x_n)\)
The symbol \(P\) is the uppercase Greek letter ‘sigma’, which is used in mathematics to denote sum.
The area under a parabola
Consider the graph of \(y = x^2\). We will find the area under the curve from \(x = 0\) to \(x = b\) using a technique due to Archimedes.
Divide the interval \([0, b]\) into \(n\) equal subintervals:
\(\left[0, \frac{b}{n}\right], \left[\frac{b}{n}, \frac{2b}{n}\right], \left[\frac{2b}{n}, \frac{3b}{n}\right], ..., \left[\frac{(n-1)b}{n}, b\right]\)
Each subinterval is the base of a rectangle with height determined by the right endpoint of the subinterval.
There is a rule for working out the sum of the first n square numbers:
\(\sum_{i=1}^n i^2 = \frac{n}{6}(n + 1)(2n + 1)\)
Area of rectangles = \(\frac{b^3}{n^3} \sum_{i=1}^n i^2 = \frac{b^3}{n^3} \times \frac{n}{6}(n + 1)(2n + 1) = \frac{b^3}{6n^2}(2n^2 + 3n + 1) = \frac{b^3}{6} \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)\)
As \(n\) becomes very large, the terms \(\frac{3}{n}\) and \(\frac{1}{n^2}\) become very small. We write:
\(\lim_{n \to \infty} \frac{b^3}{6} \left(2 + \frac{3}{n} + \frac{1}{n^2}\right) = \frac{b^3}{3}\)
We read this as: the limit of the sum as \(n\) approaches infinity is \(\frac{b^3}{3}\).
Using \(n\) left-endpoint rectangles, and considering the limit as \(n \to \infty\), also gives the area \(\frac{b^3}{3}\).
The signed area enclosed by a curve
This technique may be applied in general to a continuous function \(f\) on an interval \([a, b]\). For convenience, we will consider an increasing function.
Divide the interval \([a, b]\) into \(n\) equal subintervals.
Each subinterval is the base of a rectangle with its ‘height’ determined by the left endpoint of the subinterval.
The contribution of rectangle \(R_1\) is \((x_1 - x_0)f(x_0)\). Since \(f(x_0) < 0\), the result is negative and so we have found the signed area of \(R_1\).
The sum of the signed areas of the rectangles is \(\frac{b - a}{n} \sum_{i=0}^{n-1} f(x_i)\)
If the limit as \(n \to \infty\) exists, then we can make the following definition:
\(\int_a^b f(x) dx = \lim_{n \to \infty} \left(\frac{b - a}{n} \sum_{i=0}^{n-1} f(x_i)\right)\)
We could also have used the right-endpoint estimate: the left- and right-endpoint estimates will converge to the same limit as \(n\) approaches infinity. Definite integrals may be defined as the limit of suitable sums, and the fundamental theorem of calculus holds true under this definition.