Dove's LibraryDove's Library

6.6 - Applying Properties of Definite Integrals

Review: Area Under the Curve

The definite integral ∫abf(x) dx\int_a^b f(x) \, dx is the area under the curve from x=ax = a to x=bx = b.
Example: calculating area from a graph
6.6 - Applying Properties of Definite Integrals
Calculate ∫012f(x) dx\int_0^{12} f(x) \, dx. Add the areas above the axis, subtract the ones below.
  • Semicircle (0 to 4): πr22=π(2)22=2π\frac{\pi r^2}{2} = \frac{\pi(2)^2}{2} = 2\pi
  • Region 1 (4 to 6): area 2
  • Region 2 (6 to 8): area 2
  • Triangle (8 to 9): area 1
  • Region below the axis (9 to 12): area 4, so it counts as −4-4
∫012f(x) dx=2π+2+2+1−4=2π+1\int_0^{12} f(x) \, dx = 2\pi + 2 + 2 + 1 - 4 = 2\pi + 1

Properties of Definite Integrals

Property 1: reversal of limits
∫baf(x) dx=−∫abf(x) dx\boxed{\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx}
Switching the limits of integration flips the sign. Going “backwards” reverses the sign of the areas.
Using the function above:
∫120f(x) dx=−∫012f(x) dx=−(2π+1)=−2π−1\int_{12}^0 f(x) \, dx = -\int_0^{12} f(x) \, dx = -(2\pi + 1) = -2\pi - 1
Property 2: identical limits
∫aaf(x) dx=0\boxed{\int_a^a f(x) \, dx = 0}
No width means no area.
Property 3: constant multiple
∫abcf(x) dx=c∫abf(x) dx\boxed{\int_a^b cf(x) \, dx = c\int_a^b f(x) \, dx}
Constants can be factored out of the integral. For example:
∫153f(x) dx=3∫15f(x) dx\int_1^5 3f(x) \, dx = 3\int_1^5 f(x) \, dx
Property 4: additivity of the integration domain
∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\boxed{\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx}
where cc is any point between aa and bb. The integral can be split at any intermediate point.
∫012f(x) dx=∫04f(x) dx+∫412f(x) dx\int_0^{12} f(x) \, dx = \int_0^4 f(x) \, dx + \int_4^{12} f(x) \, dx
6.6 - Applying Properties of Definite Integrals
Property 5: sum of functions
∫ab[f(x)+g(x)] dx=∫abf(x) dx+∫abg(x) dx\boxed{\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx}
Property 6: difference of functions
∫ab[f(x)−g(x)] dx=∫abf(x) dx−∫abg(x) dx\boxed{\int_a^b [f(x) - g(x)] \, dx = \int_a^b f(x) \, dx - \int_a^b g(x) \, dx}
Integration distributes over both addition and subtraction.

Practice Using the Properties

Given: - ∫15f(x) dx=−3\int_1^5 f(x) \, dx = -3 - ∫−21f(x) dx=4\int_{-2}^1 f(x) \, dx = 4 - ∫−21g(x) dx=8\int_{-2}^1 g(x) \, dx = 8 - ∫812f(x) dx=4\int_8^{12} f(x) \, dx = 4
Problem 1: ∫76f(x) dx\int_7^6 f(x) \, dx
Reversal of limits (Property 1):
∫76f(x) dx=−∫67f(x) dx\int_7^6 f(x) \, dx = -\int_6^7 f(x) \, dx
We don’t have any information about the interval [6,7][6, 7], so we can’t get a number. But the property still tells us how the two are related.
Problem 2: ∫128−3f(x) dx\int_{12}^8 -3f(x) \, dx
Use Properties 1 and 3:
∫128−3f(x) dx=−3∫128f(x) dx=−3⋅(−∫812f(x) dx)=3∫812f(x) dx\int_{12}^8 -3f(x) \, dx = -3\int_{12}^8 f(x) \, dx = -3 \cdot \left(-\int_8^{12} f(x) \, dx\right) = 3\int_8^{12} f(x) \, dx=3(4)=12= 3(4) = 12
Problem 3: ∫51f(x) dx\int_5^1 f(x) \, dx
Reversal of limits:
∫51f(x) dx=−∫15f(x) dx=−(−3)=3\int_5^1 f(x) \, dx = -\int_1^5 f(x) \, dx = -(-3) = 3
Problem 4: ∫−25f(x) dx\int_{-2}^5 f(x) \, dx
Split the interval (Property 4):
∫−25f(x) dx=∫−21f(x) dx+∫15f(x) dx=4+(−3)=1\int_{-2}^5 f(x) \, dx = \int_{-2}^1 f(x) \, dx + \int_1^5 f(x) \, dx = 4 + (-3) = 1
Problem 5: ∫−21[f(x)+2g(x)] dx\int_{-2}^1 [f(x) + 2g(x)] \, dx
Distribute and factor out the constant (Properties 5 and 3):
=∫−21f(x) dx+2∫−21g(x) dx=4+2(8)=20= \int_{-2}^1 f(x) \, dx + 2\int_{-2}^1 g(x) \, dx = 4 + 2(8) = 20
Problem 6: ∫01f(x) dx\int_0^1 f(x) \, dx
Cannot be determined. We know ∫−21f(x) dx=4\int_{-2}^1 f(x) \, dx = 4, but that covers both [−2,0][-2, 0] and [0,1][0, 1]. Without knowing how the area is distributed between those sub-intervals, we can’t isolate [0,1][0, 1].
Problem 7: ∫1−23f(x) dx\int_1^{-2} 3f(x) \, dx
Properties 1 and 3:
∫1−23f(x) dx=3∫1−2f(x) dx=3⋅(−∫−21f(x) dx)=3⋅(−4)=−12\int_1^{-2} 3f(x) \, dx = 3\int_1^{-2} f(x) \, dx = 3 \cdot \left(-\int_{-2}^1 f(x) \, dx\right) = 3 \cdot (-4) = -12
Problem 8: ∫33[f(x)−g(x)] dx\int_3^3 [f(x) - g(x)] \, dx
Identical limits (Property 2):
∫33[f(x)−g(x)] dx=0\int_3^3 [f(x) - g(x)] \, dx = 0
No width means no area, no matter what the functions are.

Piecewise Functions

Integrate each piece separately on its own domain, then add the results.
Example
f(x)={3x<24−xx≥2f(x) = \begin{cases} 3 & x < 2 \\ 4 - x & x \geq 2 \end{cases}
6.6 - Applying Properties of Definite Integrals
Find ∫15f(x) dx\int_1^5 f(x) \, dx. Break it at x=2x = 2:
∫15f(x) dx=∫123 dx+∫25(4−x) dx\int_1^5 f(x) \, dx = \int_1^2 3 \, dx + \int_2^5 (4-x) \, dx
From 1 to 2: a rectangle of width 1 and height 3
∫123 dx=3\int_1^2 3 \, dx = 3
From 2 to 4: a triangle above the x-axis, base 2 and height 2
A1=12(2)(2)=2A_1 = \frac{1}{2}(2)(2) = 2
From 4 to 5: a triangle below the x-axis, base 1 and height 1
A2=−12(1)(1)=−0.5A_2 = -\frac{1}{2}(1)(1) = -0.5
Total:
∫15f(x) dx=3+2−0.5=4.5\int_1^5 f(x) \, dx = 3 + 2 - 0.5 = 4.5
Key point: even though there’s a jump at x=2x = 2, we still measure area from the function down to the x-axis the same way.
The same goes for a removable discontinuity (a hole, or one point moved off the graph). A single point has no width, so it adds no area and doesn’t change the integral. For example, if h(x)=2h(x) = 2 for every xx except h(3)=7h(3) = 7, then ∫05h(x) dx=2⋅5=10\int_0^5 h(x) \, dx = 2 \cdot 5 = 10, exactly the same as for the constant function y=2y = 2.

Absolute Value Functions

Absolute value functions make piecewise linear graphs, usually V-shapes.
Example: ∫05∣x−2∣ dx\int_0^5 |x - 2| \, dx
6.6 - Applying Properties of Definite Integrals
Split at the vertex x=2x = 2.
From 0 to 2: triangle with base 2, height 2
A1=12(2)(2)=2A_1 = \frac{1}{2}(2)(2) = 2
From 2 to 5: triangle with base 3, height 3
A2=12(3)(3)=4.5A_2 = \frac{1}{2}(3)(3) = 4.5
Total:
∫05∣x−2∣ dx=2+4.5=6.5\int_0^5 |x-2| \, dx = 2 + 4.5 = 6.5

Absolute Value: Inside vs. Outside the Integral

Case 1: absolute value OUTSIDE
∣∫abf(x) dx∣\left|\int_a^b f(x) \, dx\right|
  1. Calculate the integral using signed areas
  1. Take the absolute value of the result
6.6 - Applying Properties of Definite Integrals
Calculate with signs:
∫−24f(x) dx=−π2+π2+2=2\int_{-2}^4 f(x) \, dx = -\frac{\pi}{2} + \frac{\pi}{2} + 2 = 2
Take the absolute value:
∣∫−24f(x) dx∣=∣2∣=2\left|\int_{-2}^4 f(x) \, dx\right| = |2| = 2
Case 2: absolute value INSIDE
∫ab∣f(x)∣ dx\int_a^b |f(x)| \, dx
  1. All areas become positive (the negative regions flip above the axis)
  1. Add up the total positive area
Using the same function:
∫−24∣f(x)∣ dx=π2+π2+2=π+2\int_{-2}^4 |f(x)| \, dx = \frac{\pi}{2} + \frac{\pi}{2} + 2 = \pi + 2
The key difference
Type
Meaning
Negative areas?
∥∫abf(x) dx∥\left\|\int_a^b f(x) \, dx\right\|
Absolute value of the net area
Yes, they cancel normally
∫ab∥f(x)∥ dx\int_a^b \|f(x)\| \, dx
Total area, all positive
No, they all flip positive

Using a Calculator to Evaluate Integrals

For integrals that are painful by hand, the TI-84 has a built-in function.
Steps
  1. Press MATH
  1. Select option 9:fnInt(
  1. Enter fnInt(function, x, lower limit, upper limit)
  1. Press ENTER
Example 1: ∫23x−1 dx\int_2^3 \sqrt{x-1} \, dx
6.6 - Applying Properties of Definite Integrals
MATH → 9
fnInt(√(x-1), x, 2, 3)
Result: about 1.2191.219
Example 2: ∫04(4−x2) dx\int_0^4 (4-x^2) \, dx
fnInt(4-x², x, 0, 4)
Result: about −5.333-5.333
Why negative? The parabola y=4−x2y = 4 - x^2 dips below the x-axis after x=2x = 2, and on this interval the negative region outweighs the positive one.
When to use the calculator
  • Complex functions (radicals, trig, exponentials)
  • When it’s allowed on the problem
  • To check your work
  • When the geometric approach isn’t obvious
Careful: on free-response sections you may still need to show work by hand.

Key Takeaways

  • Reversal of limits: ∫baf(x) dx=−∫abf(x) dx\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx
  • Identical limits: ∫aaf(x) dx=0\int_a^a f(x) \, dx = 0
  • Constant multiple: ∫abcf(x) dx=c∫abf(x) dx\int_a^b cf(x) \, dx = c\int_a^b f(x) \, dx
  • Additivity: ∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx
  • Sum and difference: integrals distribute over addition and subtraction
  • Piecewise functions: integrate each piece on its domain, then add
  • Absolute value as piecewise: think V-shapes, and split at the vertex
  • Absolute value outside: calculate the integral first, then take the absolute value
  • Absolute value inside: all areas become positive, no negative regions
  • Calculator: TI-84 uses MATH → 9:fnInt(function, x, a, b)
  • Cannot determine: some problems just don’t give you enough information
  • dxdx means integration with respect to xx, moving along the x-axis

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.