Integration and Its Applications
FoundationIntegration completes the M2 toolkit: integration by parts and trigonometric substitution extend the basic formulas, while the definite integral, handled with the fundamental theorem of calculus, turns calculus into a tool for areas between curves and volumes of revolution.
- Integration by parts handles products of two function types: ∫u dv = uv − ∫v du, with u chosen by the priority log, inverse trig, algebraic, trig, exponential — at most twice per integral.
- Trigonometric substitution tames surds: ∫1/√(a²−x²) dx = sin⁻¹(x/a) + C and ∫1/(x²+a²) dx = (1/a)tan⁻¹(x/a) + C; mind the notation and principal values.
- Definite integrals go by the fundamental theorem: F(b) − F(a), no constant; odd functions vanish on symmetric intervals, even functions double.
文體小知識：數學概念說明文——講解分部積分法、三角代換法與定積分性質，再以兩曲線間面積與旋轉體體積展示積分的幾何應用，是 M2 微積分的收成課。
§Text Passage
1The indefinite integrals of S5 laid the foundation: the power rule, the reciprocal and exponential formulas, linearity and substitution. S6 Module 2 completes the toolkit with integration by parts and trigonometric substitution, and extends integration to the definite integral, applied to areas and volumes. Integration remains the reverse of differentiation; what changes is the depth of the tools and the breadth of the applications.
2Integration by parts handles products of two different types of functions: the integral of u with dv equals uv minus the integral of v with du. The priority for choosing u runs logarithmic, inverse trigonometric, algebraic, trigonometric, exponential — the further left, the better the choice. For example, integrating x times e to the x: take u to be x and one application gives xeˣ minus eˣ; integrating ln x also yields to the same move.
3Trigonometric substitution deals with integrals containing surds: for the square root of a squared minus x squared, set x equal to a sin θ and the root becomes a cos θ. Two standard results must be memorised: the integral of 1/√(a²−x²) is an arcsine, and the integral of 1/(x²+a²) is an arctangent times a coefficient. The curriculum also requires the notation sin⁻¹x, cos⁻¹x, tan⁻¹x and the concept of principal values.
4The definite integral turns the indefinite into a number: find the antiderivative, substitute the upper limit minus the lower — the fundamental theorem of calculus. Its properties are clear: swapping the limits changes the sign, equal limits give zero, intervals can be split, constants and sums pass through; dummy variables show the value depends only on the limits and the function. Symmetry saves time: an odd function integrates to zero over a symmetric interval, an even function to twice the half-interval.
5Geometric applications come in two themes: area and volume of revolution. For the area between two curves, solve the simultaneous equations for the intersections, take the upper function minus the lower, and integrate. For example, the region bounded by a parabola and a line is one integral away from an exact area. When a curve revolves about the x-axis it sweeps a solid of revolution, and the disc method writes the volume as the integral of π times the square of the function.
1Summary 重點摘要
Integration completes the M2 toolkit: integration by parts and trigonometric substitution extend the basic formulas, while the definite integral, handled with the fundamental theorem of calculus, turns calculus into a tool for areas between curves and volumes of revolution.
2Vocabulary cards Vocabulary
A technique for integrals of products of two function types, using ∫u dv = uv − ∫v du.
Substituting x = a sin θ and the like to remove surds such as √(a²−x²) from an integrand.
The theorem that a definite integral equals the antiderivative at the upper limit minus its value at the lower limit.
An integral with limits whose value is a number, geometrically the signed area under a curve.
The variable of a definite integral, freely renamable since the value does not depend on it.
A function with f(−x) = −f(x), integrating to zero over a symmetric interval [−a, a].
A function with f(−x) = f(x), integrating over a symmetric interval to twice the half-interval.
Slicing a solid of revolution into thin discs, giving V = π∫[f(x)]² dx.
The solid swept out when a plane curve revolves about a coordinate axis or a line parallel to one.
The prescribed range of an inverse trigonometric function, e.g. sin⁻¹x lies in [−90°, 90°].
3Step by step Steps
- Step 1: From basic integration to the M2 toolkitS5 covered the power rule, reciprocal and exponential formulas, linearity and substitution. S6 Module 2 adds three new tools: integration by parts, trigonometric substitution, and the definite integral that turns integration into numbers. For example, an integrand like x times e to the x, a product of two function types, is beyond substitution — it is exactly the territory of integration by parts; and the exact area under a parabola needs the definite integral. Build the overview first, then drill each tool — the right order for this lesson.
- Step 2: Integration by partsThe formula: the integral of u with dv equals uv minus the integral of v with du — split the product, integrate the other side. The priority for u is logarithmic, inverse trigonometric, algebraic, trigonometric, exponential: the further left, the better u. For example, for ∫xeˣ dx take u = x and dv = eˣ dx, giving xeˣ − eˣ + C; for ∫ln x dx use u = ln x, dv = dx, giving x ln x − x + C. The curriculum allows at most two applications of parts per integral; for cyclic forms like eˣ sin x, two applications bring back the original integral and solving is by transposing.
- Step 3: Trigonometric substitutionSurds have a standard substitution: for √(a²−x²) set x = a sin θ, and the root becomes a cos θ, ready for trigonometric integration. Two standard results are worth memorising: ∫1/√(a²−x²) dx = sin⁻¹(x/a) + C and ∫1/(x²+a²) dx = (1/a)tan⁻¹(x/a) + C. For example, for ∫1/√(9−x²) dx, match the standard form with a = 3 and the answer is sin⁻¹(x3) + C. The curriculum also requires the notation sin⁻¹x, cos⁻¹x, tan⁻¹x with the concept of principal values — keep your angles inside the principal range when answering.
- Step 4: The definite integral and the fundamental theoremWrite the definite integral as ∫ₐᵇ f(x) dx — the limit of signed area between the curve and the x-axis. The fundamental theorem of calculus computes it: if F′ = f, the integral equals F(b) − F(a). Properties: swapping the limits changes the sign, equal limits give zero, intervals split into two, and constants and sums pass through term by term. For example, x squared from 0 to 3: the antiderivative is x³/3, and substitution gives 9. The curriculum also requires the dummy variable: ∫ₐᵇ f(x) dx and ∫ₐᵇ f(t) dt are identical — renaming the letter changes nothing.
- Step 5: Symmetry, substitution and parts in definite integralsOn symmetric intervals, odd–even parity computes integrals instantly: an odd function with f(−x) = −f(x) integrates to zero from −a to a; an even function with f(−x) = f(x) gives twice the integral from 0 to a. For example, x cubed from −1 to 1 is zero — no antiderivative needed. Definite integrals also admit substitution and parts: with substitution, change the limits to the new variable's values; with parts, write the full uv − ∫v du expression before applying the limits. The shared pitfall of both is forgetting to change the limits, mixing old and new variables.
- Step 6: Application I — area between two curvesAreas bounded by two curves take three steps: solve the simultaneous equations for the intersection points to fix the limits; decide which curve is on top over the interval and take top minus bottom; integrate. For example, y = x² and y = x + 2 meet at x = −1 and x = 2, the line is above on the interval, and the area is ∫₋₁²[(x+2)−x²] dx = 92. If part of the curve lies below the x-axis, that strip integrates to a negative number, so split the region and add absolute values — integrating straight across cancels the areas, the biggest trap of area questions.
- Step 7: Application II — volumes of revolution (disc method)When a curve revolves about a coordinate axis or a line parallel to one, it sweeps a solid of revolution, handled by the disc method: slice the solid into thin discs whose radii are the distances from the curve to the axis of rotation and whose thickness is dx, each disc contributing π times radius squared times dx, and the limit of the sum is V = π∫ₐᵇ [f(x)]² dx. For example, y = √x from 0 to 4 about the x-axis gives V = π∫₀⁴ x dx = 8π; rotating about y = −1 instead makes the radius √x + 1, rewriting the integrand. State the disc radius first, then the integral — a complete presentation.
5Worked example Example
Example: find the area of the region bounded by the parabola y = x² and the line y = x + 2.
- Solve x² = x + 2, i.e. x² − x − 2 = 0; factorising gives (x−2)(x+1) = 0, so the curves meet at x = −1 and x = 2
- On [−1, 2] the line lies above the parabola (check with x = 0: 2 > 0), so the area = ∫₋₁² [(x+2) − x²] dx
- Antiderivative: ∫[(x+2)−x²] dx = x²/2 + 2x − x³/3
- Apply the limits: (2 + 4 − 83) − (12 − 2 + 13) = 103 − (−76) = 276 = 92
- Answer: The area is 92 square units. Three marking points: solving the simultaneous equations for the limits, identifying top and bottom and writing top minus bottom, and substituting correctly. Integrating x² − (x+2) instead gives −92 — an area cannot be negative, so a sign check intercepts the error.
6Common mistakes Mistakes
7Quick check Quick check
Q1What kind of integrand calls for integration by parts? Give one example.
Q2Evaluate ∫ x eˣ dx.
Q3Evaluate ∫₋₁¹ x³ dx.
Q4The curve y = √x on [0, 4] revolves about the x-axis. Find the volume of the solid.
Q5Evaluate ∫ 1/(x²+4) dx.
8Recap Recap
- Integration by parts handles products of two function types: ∫u dv = uv − ∫v du, with u chosen by the priority log, inverse trig, algebraic, trig, exponential — at most twice per integral.
- Trigonometric substitution tames surds: ∫1/√(a²−x²) dx = sin⁻¹(x/a) + C and ∫1/(x²+a²) dx = (1/a)tan⁻¹(x/a) + C; mind the notation and principal values.
- Definite integrals go by the fundamental theorem: F(b) − F(a), no constant; odd functions vanish on symmetric intervals, even functions double.
- Area between curves: solve for the intersections, take top minus bottom; a negative result means the order is flipped or the region needed splitting.
- Volumes of revolution use the disc method: V = π∫ₐᵇ [f(x)]² dx, the radius being the distance from the curve to the axis — rewritten when the axis is a parallel line.