3.3 Definite Integral
Definite integration extends the concept of integration by evaluating integrals between specific limits, \(x = a\) and \(x = b\). Unlike indefinite integrals, a definite integral results in a numerical value rather than an algebraic expression with an arbitrary constant \(c\). Definite integrals have vital real-life applications, such as helping engineers calculate the surface area, volume of water, and wall thickness required for hydroelectric dams like the Bakun Hydroelectric Dam in Sarawak to withstand water pressure. Mathematically, definite integrals allow us to determine the exact area under or between curves, as well as the volume of solids generated when regions are revolved fully about the \(x\)-axis or \(y\)-axis.
Learning Objectives
- Determine the value of definite integrals for algebraic functions using the Fundamental Theorem of Calculus: \(\int_{a}^{b} f(x) \, dx = [g(x) + c]_{a}^{b} = g(b) - g(a)\).
- Apply the basic characteristics and properties of definite integrals to simplify and evaluate integrals.
- Understand the relationship between the limit of the sum of areas of rectangles (Riemann sums) and the exact area under a curve.
- Calculate the area of a region bounded by a curve and the \(x\)-axis or \(y\)-axis, including regions located below the \(x\)-axis or to the left of the \(y\)-axis.
- Determine the area of a region bounded between a curve and a straight line or between two curves.
- Understand the relationship between the limit of the sum of volumes of approximating cylinders and the volume of a solid of revolution.
- Calculate the generated volume when a region is revolved through \(360^\circ\) about the \(x\)-axis or the \(y\)-axis, including volumes between two curves or a line and a curve.
Summary Guide to Definite Integrals: Characteristics, Area, and Volume of Revolution
A quick visual reference distinguishing integration along the \(x\)-axis versus the \(y\)-axis for both area and volume, highlighting sign adjustments.
Along the \(x\)-axis
Area: \(A = \int_{a}^{b} y \, dx\)
Volume: \(V = \pi \int_{a}^{b} y^2 \, dx\)
Along the \(y\)-axis
Area: \(A = \int_{a}^{b} x \, dy\)
Volume: \(V = \pi \int_{a}^{b} x^2 \, dy\)
Sign adjustment
Area below the \(x\)-axis or to the left of the \(y\)-axis requires \(|\cdot|\) (absolute value).
Positive volume
Generated volume is always positive because the radius is squared.
Short Explanation
Definite integrals, their properties, area calculations and volumes of revolution.
1. Definite Integral vs Indefinite Integral
While indefinite integration gives a family of functions \(+ c\), definite integration evaluates the antiderivative at upper and lower boundaries (\(b\) and \(a\)), subtracting \(g(a)\) from \(g(b)\) so that \(c\) disappears.
2. Properties as Calculation Shortcuts
Properties such as reversing limits \(\left(\int_a^b = -\int_b^a\right)\), splitting intervals \(\left(\int_a^c = \int_a^b + \int_b^c\right)\), and factoring constants out simplify multi-step calculations.
3. Handling Negative Definite Integrals for Area
Definite integrals calculate "net signed area". If a curve dips below the \(x\)-axis, the integral evaluates to a negative number. Because geometric area must always be positive, we take the absolute value of the integral for any region below the \(x\)-axis (or to the left of the \(y\)-axis).
4. Volume of Revolution
Rotating a 2D strip of height \(y\) and width \(\delta x\) around the \(x\)-axis generates a thin disk (cylinder) of radius \(y\), thickness \(\delta x\), and volume \(\pi y^2 \delta x\). Integrating over the interval sums an infinite number of these infinitesimal disks.
Fundamental Theorem of Calculus
\(\int_{a}^{b} f(x) \, dx = g(b) - g(a)\)where \(g'(x) = f(x)\).
| Property | Formula |
|---|---|
| Identical limits | \(\int_{a}^{a} f(x) \, dx = 0\) |
| Reversing limits | \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\) |
| Constant multiple | \(\int_{a}^{b} k f(x) \, dx = k \int_{a}^{b} f(x) \, dx\) |
| Interval addition | \(\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx\) |
| Sum and difference | \(\int_{a}^{b} [f(x) \pm g(x)] \, dx = \int_{a}^{b} f(x) \, dx \pm \int_{a}^{b} g(x) \, dx\) |
Steps for Evaluating a Definite Integral
Find the indefinite integral \(g(x) + c\).
Substitute the upper limit \(b\) into \(g(x)\).
Substitute the lower limit \(a\) into \(g(x)\).
Subtract: \(g(b) - g(a)\).
Try to Answer First
Answer in your mind, then press "Check Answer".
Evaluate \(\int_{1}^{3} (6x^2 - 4x) \, dx\).
Given that \(\int_{1}^{4} f(x) \, dx = 8\), find the value of \(\int_{4}^{1} f(x) \, dx\).
Find the generated volume, in terms of \(\pi\), when the region bounded by the curve \(y = x^2\), the \(y\)-axis, and the lines \(y = 1\) and \(y = 4\) is rotated through \(360^\circ\) about the \(y\)-axis.
Common Mistakes
- Forgetting to square the function when finding volume: writing \(\int \pi y \, dx\) instead of the correct formula \(\int \pi y^2 \, dx\).
- Squaring the difference incorrectly in volume between two curves: writing \((y_1 - y_2)^2\) instead of the correct difference of squares \((y_1^2 - y_2^2)\).
- Integrating across the \(x\)-axis in a single integral when finding total area. The sections must be split at the \(x\)-intercept.
- Mixing up variables and limits: integrating with respect to \(y\) using \(x\)-limits, or failing to express \(x\) in terms of \(y\) (\(x = g(y)\)) when revolving about the \(y\)-axis.
- Carrying the constant \(c\) unnecessarily. Definite integrals do not need \(+ c\) as it cancels out during subtraction.
Concept Misunderstandings
The value of a definite integral is always equal to the geometric area.
A definite integral can be positive, negative, or zero. Geometric area is strictly non-negative. If a region is below the \(x\)-axis, the definite integral is negative, and its absolute value must be taken to represent area.
Reversing the limits of integration has no effect on the value.
Switching the upper and lower limits reverses the sign: \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\).
Volume of revolution can be negative if the curve lies below the axis.
Because the radius is squared (\(y^2\) or \(x^2\)), the integrand is always non-negative. Generated volume is always strictly positive.
Summary
- Definite Integral Formula: \(\int_{a}^{b} f(x) \, dx = g(b) - g(a)\), where \(g'(x) = f(x)\).
- Core Characteristics: \(\int_{a}^{a} f(x) \, dx = 0\); \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\); \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\).
- Area Calculation: Between curve and \(x\)-axis: \(A = \int_{a}^{b} y \, dx\) (use \(|\cdot|\) if below the \(x\)-axis). Between curve and \(y\)-axis: \(A = \int_{a}^{b} x \, dy\) (use \(|\cdot|\) if to the left of the \(y\)-axis).
- Area between two curves: \(\int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx\) or \(\int_{c}^{d} (x_{\text{right}} - x_{\text{left}}) \, dy\).
- Volume of Revolution (\(360^\circ\)): About the \(x\)-axis: \(V = \pi \int_{a}^{b} y^2 \, dx\). About the \(y\)-axis: \(V = \pi \int_{a}^{b} x^2 \, dy\).
Short Activity
Objective: Evaluate definite integrals, apply properties, and identify area and volume formulas.
A. Objective Quiz
1 Given that \(\int_{1}^{5} f(x) \, dx = 12\) and \(\int_{1}^{3} f(x) \, dx = 4\), what is the value of \(\int_{3}^{5} f(x) \, dx\)?
2 Which of the following expressions represents the generated volume when the region bounded by \(y = f(x)\), the \(x\)-axis, and the lines \(x = a\) and \(x = b\) is revolved fully about the \(x\)-axis?
B. Fill in the Blanks
3 If the region bounded by a curve and the \(x\)-axis lies entirely below the \(x\)-axis, the value of the definite integral is negative, but the area of the region is always ________.
4 According to the characteristics of definite integrals, reversing the limits gives \(\int_{a}^{b} f(x) \, dx = -\)________.
C. Matching / Drag and Drop
Drag each expression to the correct formula or property. On a phone, tap an expression first, then tap the matching box.
