AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/6/6-6)
Unit 6 · Topic 6.6
6.6 Applying Properties of Definite Integrals
Definite integrals follow a few simple rules that let you split, combine and rearrange them. With those rules and some geometry, you can find many integrals without any antiderivative at all.
Key terms
- properties of definite integrals
- additivity
- reversing limits
- geometric area
- piecewise function
The properties
These hold for integrable functions f and g and constants k. Each one makes sense if you think of signed area.
- Zero-width interval: ∫ₐᵃ f(x) dx = 0.
- Reversing the limits: ∫ from b to a of f(x) dx = −∫ₐᵇ f(x) dx. Going backward flips the sign.
- Adjacent intervals: ∫ from a to b of f(x) dx + ∫ from b to c of f(x) dx = ∫ from a to c of f(x) dx. This holds for any order of a, b and c.
- Constant multiple: ∫ₐᵇ k·f(x) dx = k∫ₐᵇ f(x) dx.
- Sum and difference: ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx.
- Constant: ∫ₐᵇ k dx = k(b − a), the area of a rectangle.
What doesn't work
There's no product rule for integrals: ∫ f(x)g(x) dx is not ∫ f(x) dx · ∫ g(x) dx. Likewise, ∫ f(x)/g(x) dx is not a quotient of integrals. If a question gives you ∫ f and ∫ g and asks for ∫ f·g, the given information isn't enough.
Also, ∫ₐᵇ |f(x)| dx is generally not |∫ₐᵇ f(x) dx|. The first counts all area as positive; the second lets areas cancel before taking the absolute value.
Using geometry
If the graph is made of lines and circles, find the area with shape formulas and attach the sign. Two shapes come up all the time:
- y = √(r² − x²) is the upper half of a circle of radius r centered at the origin. So ∫ from −r to r of √(r² − x²) dx = ½πr², and ∫₀ʳ √(r² − x²) dx = ¼πr².
- y = |x − c| makes a V shape, so its integral is the area of two triangles.
Pieces, jumps and holes
For a piecewise function, split the integral at the break points and integrate each piece with its own formula.
A function with a jump or a removable discontinuity (a hole) can still have a definite integral. A single point has no width, so it adds no area. Split at the jump and add the areas on each side. Example: if f(x) = 2 for x < 1 and f(x) = x for x ≥ 1, then ∫₀³ f(x) dx = ∫₀¹ 2 dx + ∫₁³ x dx = 2 + 4 = 6.
Vertical asymptotes are different. If the function blows up inside the interval, the integral is improper (6.13, BC only), and these rules don't apply directly.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Combining given integrals
You're told ∫₀⁵ f(x) dx = 8, ∫₃⁵ f(x) dx = −2 and ∫₀³ g(x) dx = 4. Find (a) ∫₀³ f(x) dx, (b) ∫₃⁰ [2f(x) − 3g(x)] dx and (c) ∫₀³ [f(x) + 4] dx.
Show the solutionHide the solution
- Step 1: (a) Adjacent intervals: ∫₀³ f + ∫₃⁵ f = ∫₀⁵ f, so ∫₀³ f = 8 − (−2) = 10.
- Step 2: (b) Reverse the limits, then split: ∫₃⁰ [2f − 3g] dx = −[2∫₀³ f dx − 3∫₀³ g dx] = −[2(10) − 3(4)] = −8.
- Step 3: (c) ∫₀³ f(x) dx + ∫₀³ 4 dx = 10 + 4(3) = 22.
Answer: (a) 10 (b) −8 (c) 22
- Example 2
Integrals from geometry
Evaluate (a) ∫ from −2 to 2 of [3 + √(4 − x²)] dx and (b) ∫ from −3 to 3 of |x| dx.
Show the solutionHide the solution
- Step 1: (a) Split: ∫ from −2 to 2 of 3 dx is a rectangle, 3 × 4 = 12.
- Step 2: ∫ from −2 to 2 of √(4 − x²) dx is the area of a half circle of radius 2: ½π(2²) = 2π.
- Step 3: Total: 12 + 2π.
- Step 4: (b) The graph of |x| from −3 to 3 makes two triangles, each with base 3 and height 3. Area = 2 × ½(3)(3) = 9.
Answer: (a) 12 + 2π (b) 9
- Example 3
Trap: there's no product rule
If ∫₀² f(x) dx = 3 and ∫₀² g(x) dx = 5, can you find ∫₀² f(x)g(x) dx?
Show the solutionHide the solution
- Step 1: The integral of a product isn't the product of the integrals, so 15 is not justified.
- Step 2: Counterexample: f(x) = 1.5 and g(x) = 2.5 give ∫₀² f·g dx = 3.75·2 = 7.5, not 15.
- Step 3: Different functions with the same two given integrals can give different answers, so the information isn't enough.
Answer: No. The given information doesn't determine ∫₀² f(x)g(x) dx.
Common mistakes
- Forgetting the sign change when the limits are reversed.
- Writing ∫₀³ [f(x) + 4] dx = ∫₀³ f(x) dx + 4. The constant must be integrated too: ∫₀³ 4 dx = 12.
- Counting a semicircle as πr² instead of ½πr², or a quarter circle as ½πr².
- Assuming ∫ f·g = (∫ f)(∫ g).
On the exam
- Multiple-choice questions often give two or three integral values and ask for a combination. Draw a number line with the limits to keep track of the pieces.
- On free-response graph questions, the semicircle and triangle areas you compute here feed into accumulation-function questions (6.5).
Connected topics
Videos
Check yourself
4 questions on 6.6 Applying Properties of Definite Integrals. Pick an answer to see if you got it, and why.
Let f be the function defined by f(x) = x² + 1 for x ≤ 2 and f(x) = 5/√(1 + (x − 2)³) for x > 2. What is the value of ∫₀⁴ f(x) dx?
What is the value of ∫ from −3 to 3 of (√(9 − x²) + x³ + 2) dx?
The function f is continuous on the closed interval [0, 8]. On [0, 4], the graph of f is the lower half of the circle of radius 2 centered at (2, 0). On [4, 8], the graph of f consists of two line segments, one from (4, 0) to (6, 2) and one from (6, 2) to (8, −2).
Let g be the function defined by g(x) = ∫₀ˣ f(t) dt for 0 ≤ x ≤ 8.
Described function
What is the value of g(8)?
| x | f(x) | f′(x) | g(x) | g′(x) |
|---|---|---|---|---|
| 1 | 5 | −2 | 4 | 3 |
| 2 | 1 | 4 | 3 | −1 |
| 3 | −2 | 1 | 2 | −2 |
| 4 | 3 | 6 | 1 | −1 |
Selected values of f, f′, g and g′
The functions f and g are differentiable, and f′ and g′ are continuous for all real numbers. Selected values are given in the table. What is the value of ∫₁³ (2f′(x) + 3) dx?
0 of 4 answered