Applying Properties of Definite Integrals: Study Guide (Calculus AB 6.6)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer. ∫ (a to b) f(x) dx means the definite integral of f from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. |x| is the absolute value of x.

Integrals from geometry

In Topic 6.3 the definite integral was defined as a limit of Riemann sums. You will learn the quick way to evaluate most integrals in Topic 6.7. But when the region under the graph is made of shapes you know, you can find the integral straight away from their areas.

The rule: add the areas of the regions above the x-axis and subtract the areas of the regions below it.

Shape	Area
Rectangle, width w, height h	w · h
Triangle, base b, height h	(1/2) · b · h
Trapezoid, parallel sides p and q, width w	(1/2)(p + q) · w
Semicircle, radius r	(1/2)πr²
Quarter circle, radius r	(1/4)πr²

Three quick examples.

• ∫ (−1 to 3) (2x − 2) dx. The line crosses the axis at x = 1. From −1 to 1 it is below the axis: a triangle with base 2 and height 4, area 4, counted as −4. From 1 to 3 it is above: another triangle of area 4. So the integral is −4 + 4 = 0. A zero integral does not mean there is no region. It means the parts above and below balance.
• ∫ (−5 to 5) √(25 − x²) dx. Squaring y = √(25 − x²) gives x² + y² = 25 with y ≥ 0: the upper half of a circle of radius 5. The integral is (1/2)π(5²) = 25π/2.
• ∫ (0 to 6) |x − 4| dx. The graph is a V with its point at (4, 0). From 0 to 4 there is a triangle of base 4 and height 4 (area 8). From 4 to 6 there is a triangle of base 2 and height 2 (area 2). The integral is 10.

Recognise circles. y = √(r² − x²) is the top half of a circle of radius r centred at the origin. y = −√(r² − x²) is the bottom half. y = √(r² − (x − c)²) is the top half of a circle centred at (c, 0).

The properties of definite integrals

Each property comes from Riemann sums: it is true for every sum, so it stays true in the limit.

Property	In symbols	Why it holds
Zero width	∫ (a to a) f(x) dx = 0	A region with no width has no area
Reversed limits	∫ (b to a) f(x) dx = −∫ (a to b) f(x) dx	Going right to left makes every Δx negative
Constant multiple	∫ (a to b) k·f(x) dx = k ∫ (a to b) f(x) dx	k factors out of every term of the sum
Sum and difference	∫ (a to b) [f(x) ± g(x)] dx = ∫ (a to b) f(x) dx ± ∫ (a to b) g(x) dx	A sum of (f + g)Δx terms splits into two sums
Adjacent intervals	∫ (a to b) f(x) dx + ∫ (b to c) f(x) dx = ∫ (a to c) f(x) dx	Area from a to b plus area from b to c is area from a to c
Constant integrand	∫ (a to b) k dx = k(b − a)	A rectangle of height k and width b − a

The adjacent-interval rule works in any order. It holds even when b is not between a and c, as long as f is integrable on an interval containing all three. For example, ∫ (0 to 5) f = ∫ (0 to 8) f + ∫ (8 to 5) f. The reversed-limit rule makes the extra piece cancel correctly.

Two useful rearrangements:

• ∫ (b to c) f(x) dx = ∫ (a to c) f(x) dx − ∫ (a to b) f(x) dx
• ∫ (a to b) [f(x) + k] dx = ∫ (a to b) f(x) dx + k(b − a)

What is not a property

• Products. ∫ f(x)g(x) dx is not (∫ f(x) dx)(∫ g(x) dx). Test it: ∫ (0 to 2) x² dx = 8/3, but (∫ (0 to 2) x dx)² = 2² = 4.
• Quotients. Likewise ∫ f/g is not ∫ f divided by ∫ g.
• Absolute values. ∫ (a to b) |f(x)| dx is the total area. |∫ (a to b) f(x) dx| is the size of the net area. They are equal only when f does not change sign. For f(x) = x on [−1, 1], the first is 1 and the second is 0.
• Adding a constant. ∫ (a to b) [f(x) + 3] dx is not ∫ (a to b) f(x) dx + 3. The constant must be multiplied by the width, b − a.

Integrals of functions with jumps and holes

The definition of the definite integral also works for some functions that are not continuous.

• Removable discontinuity (a hole, or a single moved point). Changing f at one point does not change the integral. A single point is a vertical segment with zero width, so it has no area. In a Riemann sum, that one point affects at most one or two terms, and their width goes to 0.
• Jump discontinuity. Split the integral at the jump using the adjacent-interval property. Evaluate each piece using the formula that holds on that piece. The value of f exactly at the jump does not matter, for the same reason.

So a function that is continuous except for finitely many holes and jumps on [a, b], and stays bounded, has a definite integral there.

Boundary. A vertical asymptote inside [a, b] is different: the function is unbounded, so these rules do not apply. Integrals like that are improper integrals, a Calculus BC topic (Topic 6.13). In this topic, never apply the properties across a vertical asymptote.

Worked example 1: combining given values

Question. f and g are continuous, with

∫ (0 to 6) f(x) dx = 10, ∫ (4 to 6) f(x) dx = −3, ∫ (0 to 4) g(x) dx = 7.

Find: (a) ∫ (0 to 4) f(x) dx (b) ∫ (6 to 4) 5f(x) dx (c) ∫ (0 to 4) [2f(x) − 3g(x)] dx (d) ∫ (0 to 4) [f(x) + 2] dx

(a) Adjacent intervals: ∫ (0 to 4) f + ∫ (4 to 6) f = ∫ (0 to 6) f. So ∫ (0 to 4) f = 10 − (−3) = 13.

(b) Reverse the limits, then take out the constant: ∫ (6 to 4) 5f = −5 ∫ (4 to 6) f = −5 × (−3) = 15.

(c) Split the difference and take out constants: 2 ∫ (0 to 4) f − 3 ∫ (0 to 4) g = 2(13) − 3(7) = 26 − 21 = 5.

(d) ∫ (0 to 4) f + ∫ (0 to 4) 2 dx = 13 + 2(4 − 0) = 21.

Check. In (d) a common slip gives 13 + 2 = 15. The constant 2 is a height; over a width of 4 it adds a rectangle of area 8.

Worked example 2: a function with a jump and a hole

Question. Let

• h(x) = √(4 − x²) for −2 ≤ x < 0
• h(x) = x − 3 for 0 ≤ x ≤ 5, except that h(2) = 4.

(a) Find ∫ (−2 to 5) h(x) dx. (b) Find ∫ (5 to −2) 4h(x) dx. (c) Explain why the value h(2) = 4 has no effect on your answers.

(a) h has a jump at x = 0 (from height 2 on the left to −3 at x = 0) and a moved point at x = 2. Neither stops the integral. Split at the jump:

∫ (−2 to 5) h = ∫ (−2 to 0) √(4 − x²) dx + ∫ (0 to 5) (x − 3) dx

• First piece. y = √(4 − x²) on [−2, 0] is a quarter circle of radius 2. Area (1/4)π(2²) = π.
• Second piece. The line x − 3 crosses the axis at x = 3. From 0 to 3: a triangle below the axis with base 3 and height 3, area 9/2, counted as −9/2. From 3 to 5: a triangle above with base 2 and height 2, area +2.

Total: π − 9/2 + 2 = π − 5/2, about 0.64.

(b) Reverse the limits and take out the 4: ∫ (5 to −2) 4h = −4 ∫ (−2 to 5) h = −4(π − 5/2) = 10 − 4π, about −2.57.

(c) h differs from the line x − 3 at only one point, x = 2. A single point is a segment of zero width, so it adds no area. The integral of h from 0 to 5 is the same as the integral of x − 3.

Interpretation. The total area between h and the axis is π + 9/2 + 2 = π + 13/2. The integral is much smaller because the region below the axis cancels most of the rest.

Worked example 3: geometry and properties together

Question. (a) Find ∫ (−4 to 4) [2√(16 − x²) − 3] dx. (b) Find the constant k for which ∫ (0 to 4) (k − x) dx = 6.

(a) Split the difference and take out the constant:

∫ (−4 to 4) [2√(16 − x²) − 3] dx = 2 ∫ (−4 to 4) √(16 − x²) dx − ∫ (−4 to 4) 3 dx

• √(16 − x²) on [−4, 4] is the top half of a circle of radius 4: area (1/2)π(4²) = 8π.
• ∫ (−4 to 4) 3 dx is a rectangle of height 3 and width 8: area 24.

So the integral is 2(8π) − 24 = 16π − 24, about 26.3.

(b) Split: ∫ (0 to 4) k dx − ∫ (0 to 4) x dx = 4k − 8, because the first is a rectangle k by 4 and the second is a triangle with base 4 and height 4. Set 4k − 8 = 6, so k = 7/2.

Check. With k = 7/2 the line y = 7/2 − x is above the axis from 0 to 3.5 (triangle area (1/2)(3.5)(3.5) = 6.125) and below from 3.5 to 4 (area (1/2)(0.5)(0.5) = 0.125). 6.125 − 0.125 = 6. It agrees.

Common misconceptions

• “Integral = area.” A definite integral is a signed area. Regions below the axis count as negative, so an integral can be 0 or negative.
• Forgetting the width for a constant. ∫ (a to b) [f(x) + k] dx needs + k(b − a), not + k.
• Splitting a product. ∫ fg is not (∫ f)(∫ g). Only sums, differences and constant multiples split.
• Dropping the sign when limits are reversed. ∫ (6 to 4) f is the negative of ∫ (4 to 6) f.
• “A discontinuous function has no integral.” Jumps and holes are allowed. Split at a jump; ignore a single point.
• Using the wrong formula at a jump. On each piece, use the formula that holds on that piece, not the value at the endpoint.
• Applying the rules across a vertical asymptote. That is an improper integral (Calculus BC), not covered by these properties.
• Misreading circle formulas. √(r² − x²) is a semicircle of radius r, not r². The area of a semicircle is half of πr², not half of 2πr.

Where this leads

These rules turn many integrals into short calculations. In Topic 6.7, The Fundamental Theorem of Calculus and Definite Integrals, you will evaluate integrals with antiderivatives, and you will use the sum and constant-multiple rules on every line. The area calculations here are exactly what you used to find values of accumulation functions in Topic 6.5. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.