Definite Integrals from Graphs and Properties
- Evaluate a definite integral by computing signed geometric area
- Apply the additivity, reversal, and constant-multiple properties
- Combine given integral values to obtain a requested one
A definite integral is signed area
When a question supplies a graph made of line segments, semicircles, and triangles rather than a formula, evaluate the integral as signed geometric area: area above the x-axis counts positive, area below counts negative. A semicircle of radius 3 contributes (1/2)π(3²) = 4.5π; a triangle contributes half base times height; a trapezoid contributes the average of the parallel sides times the width. This is often faster than antidifferentiation even when a formula is available, and for a graph with no formula it is the only method. Watch the sign on every region — a graph designed for this question will dip below the axis specifically to test it.
The properties, and the one with a sign
Four properties do all the work of recombining given values. Additivity: ∫ₐᵇ + ∫ᵦᶜ = ∫ₐᶜ, valid for any ordering of the numbers. Reversal: ∫ᵦᵃ f = −∫ₐᵇ f, the property students forget and the one questions are built around. Zero width: ∫ₐᵃ f = 0. Linearity: ∫(cf + dg) = c∫f + d∫g, so constants come out and sums split. Note what is not a property: there is no rule for the integral of a product, so ∫fg dx cannot be broken into ∫f·∫g, and no rule for a quotient. Attempting either is a common and costly error.
Recombining given values
A standard exam item supplies two or three integral values and asks for a fourth. The method is bookkeeping: write the target as a signed combination of what you were given, using additivity and reversal. If ∫₀⁵ f = 12 and ∫₀³ f = 7, then ∫₃⁵ f = 12 − 7 = 5 by additivity. If asked for ∫₅³ f, that is −5 by reversal. If asked for ∫₀⁵ (2f(x) + 3)dx, that is 2(12) + 3(5) = 24 + 15 = 39 — the constant 3 integrates to 3 times the interval width, which is the piece most often dropped. Working symbolically before substituting numbers prevents nearly all errors here.
Given ∫₀⁴ f(x)dx = 10 and ∫₀⁷ f(x)dx = 6, find (a) ∫₄⁷ f(x)dx and (b) ∫₀⁴ [3f(x) − 2]dx.
- 1.For (a), use additivity: ∫₀⁴ f + ∫₄⁷ f = ∫₀⁷ f, so 10 + ∫₄⁷ f = 6.
- 2.Therefore ∫₄⁷ f = 6 − 10 = −4. The negative value tells you f spends more area below the axis than above it on [4, 7].
- 3.For (b), split by linearity: ∫₀⁴ [3f(x) − 2]dx = 3∫₀⁴ f(x)dx − ∫₀⁴ 2 dx.
- 4.The first term is 3(10) = 30.
- 5.The second is 2 times the interval width: 2(4 − 0) = 8.
- 6.So the answer is 30 − 8 = 22.
There is no product rule for integrals. ∫f(x)g(x)dx is not ∫f·∫g, and no manipulation makes it so. If a question appears to require that, you have misread it or the intended method is substitution.
The graph of f on [0, 6] consists of a triangle above the axis on [0, 2] with height 4, and a triangle below the axis on [2, 6] with depth 3. What is ∫₀⁶ f(x)dx?
If ∫₂⁸ f(x)dx = 15, then ∫₈² f(x)dx equals
Answer the 2 checkpoints as you read.
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