AOS3 Topic 4: 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 \( h f(d) \leq A(x + h) - A(x) \leq h f(c) \).

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 \( f \).

Properties of the Definite Integral

• \[ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \]
• \[ \int_a^a f(x) \, dx = 0 \]
• \[ \int_a^b k f(x) \, dx = k \int_a^b f(x) \, dx \]
• \[ \int_a^b \left( f(x) \pm g(x) \right) \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx \]
• \[ \int_a^b f(x) \, dx = - \int_b^a f(x) \, dx \]

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 \(\sum\) 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: \([0, \frac{b}{n}], [\frac{b}{n}, \frac{2b}{n}], [\frac{2b}{n}, \frac{3b}{n}], \ldots, [\frac{(n-1)b}{n}, b]\).

Each subinterval is the base of a rectangle with height determined by the right endpoint of the subinterval.

Area of rectangles = \(\frac{b}{n}\left(\left(\frac{b}{n}\right)^2 + \left(\frac{2b}{n}\right)^2 + \left(\frac{3b}{n}\right)^2 + \cdots + \left(\frac{nb}{n}\right)^2\right)\)

= \(\frac{b}{n} \left(\frac{b^2}{n^2} + \frac{4b^2}{n^2} + \frac{9b^2}{n^2} + \cdots + \frac{n^2b^2}{n^2}\right)\)

= \(\frac{b^3}{n^3}\left(1 + 4 + 9 + \cdots + n^2\right)\)

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 \(\frac{b^3}{3}\).

Finding the Area Under a Curve

Recall that the definite integral \( \int_a^b f(x) \, dx \) gives the net signed area ‘under’ the curve.

Finding the Area of a Region

• If \( f(x) \geq 0 \) for all \( x \in [a, b] \), then the area \( A \) of the region contained between the curve, the x-axis, and the lines \( x = a \) and \( x = b \) is given by: \[ A = \int_a^b f(x) \, dx \]
• If \( f(x) \leq 0 \) for all \( x \in [a, b] \), then the area \( A \) of the region contained between the curve, the x-axis, and the lines \( x = a \) and \( x = b \) is given by: \[ A = -\int_a^b f(x) \, dx = \int_b^a f(x) \, dx \]
• If \( c \in (a, b) \) with \( f(c) = 0 \) and \( f(x) \geq 0 \) for \( x \in (c, b] \) and \( f(x) \leq 0 \) for \( x \in [a, c) \), then the area \( A \) of the shaded region is given by: \[ A = \int_c^b f(x) \, dx + \left( - \int_a^c f(x) \, dx \right) \]

Note: In determining the area ‘under’ a curve \( y = f(x) \), the sign of \( f(x) \) in the given interval is the critical factor.