Difficulty: Introductory to Intermediate | Prerequisites: Basic integration (antiderivatives), sigma notation helpful but not required.
The properties of definite integrals are the algebraic rules that let you break apart, combine and rescale integrals without ever computing an antiderivative. They sit right between learning what a definite integral is (the limit of Riemann sums) and learning how to evaluate one (the Fundamental Theorem of Calculus). If you have been away for a few lectures, the one thing to know is that a definite integral behaves a lot like a sum: you can pull constants out, split it across addition, and chop the interval into pieces.
Definite integrals obey a handful of clean algebraic rules: you can factor out constants, split sums and differences, glue adjacent intervals together, and reverse limits by flipping the sign. Mastering these properties means you can simplify ugly-looking integrals into pieces you already know, which is exactly what exam problems test.
Definite integral
The integral of f(x) from a to b, written ∫ₐᵇ f(x) dx, representing the net signed area between the curve y = f(x) and the x-axis over the interval [a, b]. Think of it as the total accumulated value, where area below the x-axis counts as negative.
Linearity of integration
The principle that integration distributes over addition and scalar multiplication. In simple terms, you can break a complicated integral into smaller, friendlier pieces and handle each one separately.
Constant multiple rule (for integrals)
The rule stating ∫ₐᵇ c · f(x) dx = c · ∫ₐᵇ f(x) dx for any constant c. In plain English, you can pull a constant factor outside the integral sign, just as you would with a sum.
Sum and difference rule (for integrals)
The rule stating ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx. In simple terms, the integral of a sum equals the sum of the integrals, and likewise for differences.
Additivity over intervals (interval splitting)
The property ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx. Think of it as gluing two adjacent stretches of area together into one longer stretch.
Reversal of limits
The rule ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx. Swapping the upper and lower limits flips the sign, nothing more.
Net signed area
The value a definite integral returns: area above the x-axis is positive, area below is negative, and the integral gives the combined total. This is not the same as total (unsigned) area.
If c is any real constant: ∫ₐᵇ c · f(x) dx = c · ∫ₐᵇ f(x) dx
You simply "pull c out" of the integral.
Example: given ∫₁⁵ f(x) dx = 8, then ∫₁⁵ 3f(x) dx = 3 · 8 = 24.
∫ₐᵇ [f(x) + g(x)] dx = ∫ₐᵇ f(x) dx + ∫ₐᵇ g(x) dx
∫ₐᵇ [f(x) − g(x)] dx = ∫ₐᵇ f(x) dx − ∫ₐᵇ g(x) dx
You break the integral apart across the plus or minus sign.
Example: given ∫₁⁵ f(x) dx = 8 and ∫₁⁵ g(x) dx = −3, then ∫₁⁵ [f(x) + g(x)] dx = 8 + (−3) = 5.
The two rules above work together. Pull out the constants first, then combine.
Example: ∫₁⁵ [3f(x) − 4g(x)] dx = 3 · ∫₁⁵ f(x) dx − 4 · ∫₁⁵ g(x) dx = 3(8) − 4(−3) = 24 + 12 = 36.
Example: ∫₁⁵ [f(x) + ½g(x)] dx = 8 + ½(−3) = 8 − 3/2 = 13/2.
∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫ₐᶜ f(x) dx
Adjacent integrals of the same function combine into one integral spanning the full interval.
This also works in reverse: one integral can be split into two at any interior point.
Example: ∫₀² f(x) dx + ∫₂⁵ f(x) dx = ∫₀⁵ f(x) dx.
Example: ∫₂⁹ f(x) dx − ∫₂⁶ f(x) dx = ∫₆⁹ f(x) dx (subtracting the overlapping part isolates the remainder).
∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
Swapping the limits of integration negates the value.
This is essential when you encounter integrals whose limits run "backwards" (upper limit smaller than lower limit).
Example: ∫₅¹ f(x) dx + ∫₈⁵ f(x) dx: rewriting ∫₅¹ as −∫₁⁵ and ∫₈⁵ as −∫₅⁸ may help you recombine.
Problems 9 and 10 on the worksheet use variable endpoints (c, d, k, h) to test whether you can apply the same rules without specific numbers.
∫_c^d f(x) dx + ∫_k^c f(x) dx: reorder using additivity to get ∫_k^d f(x) dx (since the c endpoints cancel).
∫_c^(d+h) f(x) dx − ∫_c^h f(x) dx: use splitting to isolate ∫_h^(d+h) f(x) dx.
\int_a^b c \cdot f(x)\,dx = c \int_a^b f(x)\,dx\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\int_a^a f(x)\,dx = 0These properties are how engineers and scientists decompose complicated integrals in practice. If you need the total energy delivered over a 10-hour window but you only have data for the first 6 hours and the last 4 separately, additivity over intervals is what lets you add the two pieces.
Students often try to pull a variable expression (like x or f(x)) outside the integral the way they pull a constant. Only true constants come out.
Students sometimes forget that ∫ₐᵇ g(x) dx can be negative. A negative integral value means the net area is below the x-axis, not that something has gone wrong.
Confusing additivity with the product rule: ∫ₐᵇ f(x) · g(x) dx does NOT equal ∫ₐᵇ f(x) dx · ∫ₐᵇ g(x) dx. Integration does not distribute over multiplication.
Reversing limits and forgetting the sign change: ∫₅¹ f(x) dx is not the same as ∫₁⁵ f(x) dx. You must negate.
⚠️ Expect problems that give you two or three known integral values and ask you to find a new integral by applying linearity. Problems 1 to 4 on this worksheet are the template.
⚠️ Interval-splitting problems (like problems 5 to 10) appear frequently. You need to spot which endpoints cancel and which remain.
⚠️ Problems with variable endpoints (letters instead of numbers) test whether you understand the properties conceptually, not just mechanically.
True or False: ∫ₐᵇ 5f(x) dx = 5 ∫ₐᵇ f(x) dx. (True)
True or False: ∫ₐᵇ f(x) · g(x) dx = ∫ₐᵇ f(x) dx · ∫ₐᵇ g(x) dx. (False)
Fill in the blank: ∫₃⁷ f(x) dx + ∫₇¹⁰ f(x) dx = ∫₃___ f(x) dx. (10)
True or False: ∫₅² f(x) dx = ∫₂⁵ f(x) dx. (False, it equals the negative)
Fill in the blank: If ∫₁⁵ f(x) dx = 8, then ∫₁⁵ [2f(x)] dx = ___. (16)
Q: Given ∫₁⁵ f(x) dx = 8 and ∫₁⁵ g(x) dx = −3, find ∫₁⁵ [3f(x) − 4g(x)] dx.
A: Apply linearity: 3(8) − 4(−3) = 24 + 12 = 36.
Q: Express ∫₀² f(x) dx + ∫₂⁵ f(x) dx as a single integral.
A: By additivity over intervals, ∫₀⁵ f(x) dx.
Q: Express ∫₃⁸ f(x) dx − ∫₅⁸ f(x) dx as a single integral.
A: The interval [5, 8] is removed from [3, 8], leaving ∫₃⁵ f(x) dx.
Q: Express ∫₂⁹ f(x) dx − ∫₂⁶ f(x) dx as a single integral.
A: Removing [2, 6] from [2, 9] leaves ∫₆⁹ f(x) dx.
Q: If ∫₁⁵ f(x) dx = 8 and ∫₁⁵ g(x) dx = −3, find ∫₁⁵ [f(x) + ½g(x)] dx.
A: 8 + ½(−3) = 8 − 3/2 = 13/2.
Q: Express ∫_c^d f(x) dx + ∫_k^c f(x) dx as a single integral.
A: By additivity (the c endpoints match), ∫_k^d f(x) dx.
These properties connect directly to the Fundamental Theorem of Calculus, which gives you a way to evaluate definite integrals once you have an antiderivative. Knowing the properties first means you can simplify before you integrate.
Additivity over intervals reappears in improper integrals, where you split at a point of discontinuity and evaluate each piece separately.
The linearity rules carry over into applications such as computing net change, total displacement and average value of a function.
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