AOS3 Topic 10: Integration

Integration is a fundamental concept in calculus that refers to the process of finding the integral of a function. In a geometric sense, integration is used to calculate the area under a curve defined by a function over a given interval on the x-axis.

We define an area function \( A(x) \) for a given function \( f(x) \) over an interval \([a, b]\) as follows:

\[ A(x) = \int_{a}^{x} f(t) \, dt \]

Here, \( A(x) \) represents the area under the curve of \( f(t) \) from the point \( a \) to any point \( x \) within the interval \([a, b]\).

Estimating the area under a graph

Consider the graph of a function \( f(x) \). We want to find the area under the graph. For now, we’ll assume that the graph \( y = f(x) \) is always above the x-axis, and we will estimate the area between the graph \( y = f(x) \) and the x-axis. We set left and right endpoints and estimate the area between those endpoints.

The Left-Endpoint Method

The left-endpoint method is a numerical technique used to approximate the area under a curve, specifically between two points on the x-axis. This method involves dividing the region under the curve into a series of rectangles. The height of each rectangle is determined by the value of the function at the left endpoint of each subinterval.

The right-endpoint method

The right-endpoint method is a numerical technique used to approximate the area under a curve by dividing the region under the curve into rectangles, where the height of each rectangle is determined by the value of the function at the right endpoint of each subinterval.

The right-endpoint method is a numerical technique used to approximate the area under a curve by dividing the region under the curve into rectangles, where the height of each rectangle is determined by the value of the function at the right endpoint of each subinterval.

For \( f(x) \) decreasing over \([a, b]\): Left-endpoint estimate \( \geq \) true area \( \geq \) right-endpoint estimate

For \( f(x) \) increasing over \([a, b]\): Left-endpoint estimate \( \leq \) true area \( \leq \) right-endpoint estimate

The trapezium rule

Areas of trapeziums:

• \( T_1 = \frac{1}{2} \left( f(2.0) + f(2.5) \right) \times 0.5 = 4.24375 \)
• \( T_2 = \frac{1}{2} \left( f(2.5) + f(3.0) \right) \times 0.5 = 4.11875 \)
• \( T_3 = \frac{1}{2} \left( f(3.0) + f(3.5) \right) \times 0.5 = 3.96875 \)
• \( T_4 = \frac{1}{2} \left( f(3.5) + f(4.0) \right) \times 0.5 = 3.79375 \)
• \( T_5 = \frac{1}{2} \left( f(4.0) + f(4.5) \right) \times 0.5 = 3.59375 \)
• \( T_6 = \frac{1}{2} \left( f(4.5) + f(5.0) \right) \times 0.5 = 3.36875 \)

The sum of the areas is 23.0875 square units.

Function: \( f(x) = 9 - 0.1x^2 \)

This is called the trapezium estimate for the total area.

We can see that the trapezium estimate for this example can also be calculated as:

\[ \frac{1}{2} \times h \times \left[ f(2) + 2 \cdot f(2.5) + 2 \cdot f(3) + 2 \cdot f(3.5) + 2 \cdot f(4) + 2 \cdot f(4.5) + f(5) \right] \times 0.5 \]

The trapezium estimate is the average of the left-endpoint and right-endpoint estimates.

The definite integral:

Suppose that \( f \) is a continuous function on a closed interval \([a, b]\) and that \( f(x) \) is positive for all \( x \) in this interval. Then the area under the graph of \( y = f(x) \) from \( x = a \) to \( x = b \) is called the definite integral of \( f(x) \) from \( x = a \) to \( x = b \), and is denoted by:

\[ \int_{a}^{b} f(x) \, dx \]

The function \( f \) is called the integrand, and \( a \) and \( b \) are the lower and upper limits of the integral.

By using summation notation, this limiting process can be expressed as:

\[ \int_{a}^{b} f(x) \, dx = \lim_{\Delta x \to 0} \sum_{i=1}^{n} f(x_i^*) \Delta x_i \]

where the interval \([a, b]\) is partitioned into \( n \) subintervals, with the \( i \)-th subinterval of length \( \Delta x_i \) and containing \( x_i^* \), and \( \Delta x = \max\{\Delta x_i : i = 1, 2, \dots, n \}\).

The trapezium rule can be used to obtain an approximation for the definite integral. For Example 1, we can describe this approximation for the definite integral as:

\[ \int_0^3 f(x) \, dx \approx \frac{1}{4} \left( f(0) + 2f(0.5) + 2f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + f(3) \right) \]

In general:

\[ \int_{a}^{b} f(x) \, dx \approx \frac{b - a}{2n} \left( f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n) \right) \]

For a linear function or a piecewise-defined function with linear components, the area under the graph may be found using geometric techniques.

Antidifferentiation: Indefinite Integrals

The derivative of \(x^2\) with respect to \(x\) is \(2x\). Conversely, given that an unknown expression has a derivative of \(2x\), it is clear that the unknown expression could be \(x^2\). The process of finding a function from its derivative is called antidifferentiation.

Now consider the functions \(f(x) = x^2 + 1\) and \(g(x) = x^2 - 7\).

We have \(f'(x) = 2x\) and \(g'(x) = 2x\). So, two different functions have the same derivative function.

Both \(x^2 + 1\) and \(x^2 - 7\) are said to be antiderivatives of \(2x\). If two functions have the same derivative function, then they differ by a constant. So the graphs of the two functions can be obtained from each other by translation parallel to the y-axis.

Notation

The general antiderivative of \(2x\) is \(x^2 + c\), where \(c\) is an arbitrary real number. We use the notation of Leibniz to state this with symbols:

\[\int 2x \, dx = x^2 + c\]

This is read as "the general antiderivative of \(2x\) with respect to \(x\) is equal to \(x^2 + c\)" or as "the indefinite integral of \(2x\) with respect to \(x\) is \(x^2 + c\)." To be more precise, the indefinite integral is the set of all antiderivatives, and to emphasize this, we could write:

\[\int 2x \, dx = \{ f(x) : f'(x) = 2x \} = \{ x^2 + c : c \in \mathbb{R} \}\]

This set notation is not commonly used, but it should be clearly understood that there is not a unique antiderivative for a given function. We will not use this set notation, but it is advisable to keep it in mind when considering further results.

The reason why the symbol is the same as that used for the definite integral will become evident in later sections.

In general: If \(F'(x) = f(x)\), then \[\int f(x) \, dx = F(x) + c\], where \(c\) is an arbitrary real number.

The antiderivative of \(x^r\) where \(r \neq -1\)

We know that:

\[f(x) = x^3 \implies f'(x) = 3x^2\]

\[f(x) = x^8 \implies f'(x) = 8x^7\]

\[f(x) = x^{\frac{3}{2}} \implies f'(x) = \frac{3}{2}x^{\frac{1}{2}}\]

\[f(x) = x^{-4} \implies f'(x) = -4x^{-5}\]

Reversing this process gives:

\[\int 3x^2 \, dx = x^3 + c\]

\[\int 8x^7 \, dx = x^8 + c\]

\[\int \frac{3}{2}x^{\frac{1}{2}} \, dx = x^{\frac{3}{2}} + c\]

\[\int -4x^{-5} \, dx = x^{-4} + c\]

We also have:

\[\int x^2 \, dx = \frac{1}{3}x^3 + c\]

\[\int x^{\frac{1}{2}} \, dx = \frac{2}{3}x^{\frac{3}{2}} + c\]

\[\int x^{-5} \, dx = -\frac{1}{4}x^{-4} + c\]

Generalizing:

\[\int x^r \, dx = \frac{x^{r+1}}{r+1} + c\], where \(r \in \mathbb{Q} \setminus \{-1\}\)

Note: This result can only be applied for suitable values of \(x\) for a given value of \(r\). For example, if \(r = \frac{1}{2}\), then \(x \in \mathbb{R}^{+}\) is a suitable restriction. If \(r = -2\), we can take \(x \in \mathbb{R} \setminus \{0\}\), and if \(r = 3\), we can take \(x \in \mathbb{R}\).

We also record the following results, which follow immediately from the corresponding results for differentiation:

• Sum: \[\int \left( f(x) + g(x) \right) \, dx = \int f(x) \, dx + \int g(x) \, dx\]
• Difference: \[\int \left( f(x) - g(x) \right) \, dx = \int f(x) \, dx - \int g(x) \, dx\]
• Multiple: \[\int k f(x) \, dx = k \int f(x) \, dx\], where \(k\) is a real number