Integration and Estimation (Module 1)

§Text Passage

1Summary 重點摘要

Module 1 integration is application-driven: evaluate definite integrals with the basic formulas and substitution, find areas between a curve and the x-axis, and when no antiderivative can be found, estimate integrals with the trapezoidal rule, using the second derivative's concavity to decide whether the estimate is over or under.

2Vocabulary cards Vocabulary

3Step by step Steps

1. Step 1: The M1 integration toolkit and the syllabus boundaryThe Module 1 formula table: the integral of a constant k is kx + C; x to the n integrates to x to the n+1 over n+1; one over x gives ln x + C; e to the x is its own integral. Together with linearity and substitution. Memorise the boundary: M1 excludes integration by parts and trigonometric substitution, so a form like x times e to the x needs no struggle. For example, the integral of three x squared: factor out the constant, apply the power rule, and get x cubed plus C. Master the toolkit first, then spend the effort on applications.
2. Step 2: Integration by substitution, M1 styleSubstitution handles composite forms: set the inner function as u, find du, rewrite everything in terms of u, integrate and substitute back. For example, ∫2x/(x²+1) dx with u = x²+1 and du = 2x dx becomes ∫du/u = ln(x²+1) + C. In definite integrals, change the limits to the u-values as well and no back-substitution is needed. The test: somewhere in the integrand a piece is a constant multiple of the derivative of another piece — the numerator matching the denominator's derivative, for instance. M1 substitution questions use polynomial and exponential composites; trigonometric substitution never appears.
3. Step 3: The fundamental theorem and properties of the definite integralThe definite integral is defined as the limit of sums of rectangular strips and computed by the fundamental theorem: ∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f. Properties: swapping the limits changes the sign, equal limits give zero, intervals split, constants and sums pass through term by term. For example, e to the x from 0 to 1 gives e − 1; x squared from 1 to 3 gives (27−1)/3 = 263. The answer is a number — no constant of integration; and the dummy variable concept, explicitly required, says replacing x by t changes nothing.
4. Step 4: Area questions and the signThe standard M1 area question asks for the area between a curve and the x-axis: solve f(x) = 0 for the intersection points and split the interval; strips below the axis integrate to negative numbers, so take absolute values before adding. For example, for y = x² − 4 between −3 and 3, the curve is below the axis from −2 to 2, so that piece's integral is made positive before adding the two positive flanks. Remember the boundary at the same time: M1 does not require areas between a curve and the y-axis or between two curves — do not over-revise; save the time for applications.
5. Step 5: Application questions with definite integralsM1 applications prize the argument format. Two recurring situations: recovering a total from a rate — given a marginal cost, integrate and pin the constant with an initial condition; and accumulating a quantity from a flow — given a rate over time, the total is the definite integral. Answer in five steps: define the variables, state the function and its domain, integrate, apply the condition, conclude in words. For example, a marginal cost of six x plus two over output from 0 to 10 accumulates by definite integration to 320; a missing domain or conclusion sentence is the most common lost mark.
6. Step 6: The trapezoidal ruleWhen the antiderivative cannot be found or only tabulated data exist, estimate the integral with the trapezoidal rule: split [a, b] into n equal parts of width h = (b−a)/n, approximate each strip by a trapezium, and T = (h2)[first + last + 2×(sum of the middle values)]. For example, estimate ∫₀² x² dx with two equal strips: h = 1, the y-values are 0, 1, 4, and T = (12)(0 + 4 + 2×1) = 3, slightly above the exact 83 ≈ 2.67. Finer divisions are more accurate; for data tables, feed the table values into the same formula — and remember to double the middle values.
7. Step 7: Over or under: judging with concavityWhether the trapezoidal estimate is too high or too low is decided, as the curriculum requires, by the second derivative and concavity: f''(x) > 0 means the curve opens upward, each chord lies above the curve, and the estimate is an overestimate; f''(x) < 0 puts the chords below and the estimate is an underestimate. For example, f(x) = e⁻ˣ has second derivative e⁻ˣ > 0, opening upward, so the trapezoidal estimate of ∫₀¹ e⁻ˣ dx must be high; the exact 1 − e⁻¹ ≈ 0.632 against the estimate 0.635 confirms it. Fix the order of the answer: differentiate twice, state the concavity, conclude — omit any step and marks fall.

5Worked example Example

6Common mistakes Mistakes

7Quick check Quick check

8Recap Recap

• The M1 toolkit: basic formulas (k, xⁿ, 1/x, eˣ) plus linearity and substitution; no integration by parts, no trigonometric substitution.
• Definite integrals go by the fundamental theorem: F(b) − F(a); swapping limits flips the sign, intervals split, constants and sums pass through.
• Area questions split at the intersections and take absolute values below the axis; M1 never asks for areas against the y-axis or between two curves.
• The trapezoidal rule: T = (h2)[first + last + 2×(middle sum)]; tabulated data go straight in, with middle values doubled.
• Judge over or under with the second derivative: f'' > 0 opens upward and overestimates, f'' < 0 underestimates; argue in three steps — differentiate, state concavity, conclude.