Definite Integral as Limit

1. Area Under a Curve as a Limit of a Sum

1.1. Informal Consideration

The definite integral \(\int_a^b f(x) \, dx\) can be informally understood as the area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\). A more rigorous definition involves considering this area as the limit of a sum.

1.2. Partitioning the Interval

1. Divide the interval \([a, b]\) into \(n\) subintervals. These subintervals do not necessarily need to be of equal width.
2. Let \(\delta x_i\) be the width of the \(i\)-th subinterval.
3. Choose a point \(x_i^*\) within each subinterval.

1.3. Riemann Sum

Form the Riemann sum:

This represents the sum of the areas of \(n\) rectangles, where the height of each rectangle is given by \(f(x_i^*)\) and the width is \(\delta x_i\).

1.4. Definition of the Definite Integral

The definite integral is defined as the limit of the Riemann sum as the width of the widest subinterval (denoted by \(\delta x = \max\{\delta x_i\}\)) approaches zero:

• If this limit exists, we say that \(f\) is integrable on \([a, b]\).

1.5. Equal Subintervals

If the subintervals are of equal width, then \(\delta x_i = \Delta x = \frac{b-a}{n}\) for all \(i\). The Riemann sum simplifies to:

And the definite integral becomes:

KEY TAKEAWAY: The definite integral is formally defined as the limit of a Riemann sum, representing the area under a curve as the sum of infinitely many infinitesimally thin rectangles.

2. Approximating Definite Integrals Using the Trapezium Rule

2.1. Trapezium Rule Formula

The trapezium rule provides an approximation of the definite integral using trapeziums instead of rectangles. Divide the interval \([a, b]\) into \(n\) equal subintervals of width \(h = \frac{b-a}{n}\). The trapezium rule is given by:

where \(x_i = a + ih\) for \(i = 0, 1, 2, \dots, n\).

2.2. Alternative Form

2.3. Geometric Interpretation

Each term in the sum represents the area of a trapezium with parallel sides \(f(x_{i-1})\) and \(f(x_i)\), and height \(h\).

2.4. Accuracy

• The accuracy of the trapezium rule increases as \(n\) (the number of trapeziums) increases.
• For concave-up functions, the trapezium rule overestimates the area.
• For concave-down functions, the trapezium rule underestimates the area.

2.5. Example

Approximate \(\int_0^3 (x^2 + 2) \, dx\) using the trapezium rule with \(n = 3\).

1. \(a = 0\), \(b = 3\), \(n = 3\), so \(h = \frac{3-0}{3} = 1\).
2. \(x_0 = 0\), \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\).
3. \(f(x_0) = 0^2 + 2 = 2\)
4. \(f(x_1) = 1^2 + 2 = 3\)
5. \(f(x_2) = 2^2 + 2 = 6\)
6. \(f(x_3) = 3^2 + 2 = 11\)
7. Applying the trapezium rule:

The exact value is \(\int_0^3 (x^2 + 2) \, dx = \left[ \frac{x^3}{3} + 2x \right]_0^3 = \frac{3^3}{3} + 2(3) = 9 + 6 = 15\).

In this case, the trapezium rule overestimates the true value by \(0.5\).

EXAM TIP: Pay close attention to the number of strips (\(n\)) specified in the problem. Ensure you calculate the width of each trapezium (\(h\)) correctly.

3. Signed Area

3.1. Areas Below the x-axis

When a function \(f(x)\) takes on negative values within the interval \([a, b]\), the definite integral represents the signed area.

• Areas above the x-axis contribute positively to the integral.
• Areas below the x-axis contribute negatively to the integral.

3.2. Calculating Total Area

To find the total area between the curve and the x-axis, you need to consider the intervals where the function is positive and negative separately, and take the absolute value of the integrals over the intervals where the function is negative.

Total Area \(= \int_a^b |f(x)| \, dx\)

This often involves finding the x-intercepts of the function within the interval \([a, b]\) and splitting the integral into multiple integrals:

Total Area = \(\left| \int_a^{x_1} f(x) \, dx \right| + \left| \int_{x_1}^{x_2} f(x) \, dx \right| + ...\)

where \(x_1, x_2, ...\) are the x-intercepts of \(f(x)\) in \([a, b]\).

3.3 Trapezium Rule and Signed Area

The trapezium rule can be applied directly even when the function takes on negative values. The result will be an approximation of the signed area. To approximate the total area, you must first determine where the function crosses the x-axis, and then apply the trapezium rule to each interval separately, taking the absolute value of the results for intervals where the function is negative.

COMMON MISTAKE: Forgetting to take the absolute value of the integral when calculating the total area when the function is negative.

4. Fundamental Theorem of Calculus

4.1. Statement

If \(f\) is a continuous function on an interval \([a, b]\), then

where \(F(x)\) is any antiderivative of \(f(x)\), i.e., \(F'(x) = f(x)\).

4.2. Significance

The Fundamental Theorem of Calculus links differentiation and integration. It provides a method for evaluating definite integrals using antiderivatives.

STUDY HINT: Practice finding antiderivatives of various functions. This is crucial for evaluating definite integrals using the Fundamental Theorem of Calculus.

5. Examples of Limits of Sums

5.1. Area under \(y = x^2\)

Consider the area under the curve \(y = x^2\) from \(x = 0\) to \(x = b\). Dividing the interval \([0, b]\) into \(n\) equal subintervals of width \(\Delta x = \frac{b}{n}\), and using right endpoints, the Riemann sum is:

Using the formula \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\), we have:

Taking the limit as \(n \to \infty\):

Therefore, \(\int_0^b x^2 \, dx = \frac{b^3}{3}\).

VCAA FOCUS: VCAA exam questions often involve approximating definite integrals using the trapezium rule and comparing the approximation to the exact value obtained using the Fundamental Theorem of Calculus.