AQA A-level Mathematics 7357, section H (Integration), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: differentiation from section G. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
A runner's velocity graph dips below the time axis for a while, then climbs back. The "area under the curve" is nearly zero — did the runner go nowhere?
Start with a prediction. State what a signed area means, and how total distance differs from displacement.
Build the model
Integration reverses differentiation: an antiderivative 原函数 of axⁿ is axⁿ⁺¹/(n+1) + C for n ≠ −1, while 1/x integrates to ln|x| + C. A definite integral 定积分 from a to b is signed accumulation: F(b) − F(a).
Signed area cancels where the curve crosses the axis, so total positive area needs the integral SPLIT at each crossing. The Fundamental Theorem 微积分基本定理 connects both directions: differentiating an accumulation function returns the integrand.
antiderivative: A function whose derivative is the given integrand; definite integral: The signed accumulation of a function between two limits; Fundamental Theorem: The two-way link that differentiating the accumulation function gives back the integrand.
Choose evidence that can test it
Verify every antiderivative by differentiating it back, and keep the constant C for indefinite integrals — a supplied point fixes it. The n = −1 case is separate: 1/x integrates to ln|x|, never the power rule with division by zero.
Before computing total area, find the sign changes of the integrand on the interval and split there; a velocity with a sign change makes displacement and total distance different answers.
Work from known quantities
State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
Known: v(t) = 3t² − 3 for 0 ≤ t ≤ 2. An antiderivative is t³ − 3t. Displacement = [t³ − 3t] from 0 to 2 = (8 − 6) − 0 = 2 m. But v = 0 at t = 1, so total distance = −[t³ − 3t] from 0 to 1 + [t³ − 3t] from 1 to 2 = 2 + 4 = 6 m.
Example:
Find the area between y = x² and the x-axis from 0 to 1, then the right-end rectangle estimate with 4 strips, using the same sequence. (Answer: exact ∫₀¹ x² dx = 1/3 ≈ 0.333; the estimate is (1/4)(1/16 + 1/4 + 9/16 + 1) = 15/32 ≈ 0.469, an overestimate for this increasing convex curve.)
Check the conclusion and its limits
Integration is the limit of rectangle sums; a finite strip count only estimates it, and the error depends on curvature. Do not integrate across a discontinuity such as 1/x at x = 0, and remember signed area can be zero while total area is positive.
Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
A definite integral always equals the total positive area between the curve and the axis. This claim is false: the integral is SIGNED — regions below the axis count negatively, so displacement can be small or zero while the total distance travelled is large.
Key:
Integration: antiderivatives verified by differentiating back, with C fixed by a supplied point; definite integrals are signed accumulation F(b) − F(a); total area needs a split at every sign change, and the Fundamental Theorem links accumulation to the integrand.
