Properties of the Definite Integral

Because the definite integral is a limit of sums, it inherits the algebra of sums: you can split integrands, pull out constants, chop the interval into adjacent pieces, and run it backwards — each move has a simple, exact rule.

By the end you'll be able to combine linearity, additivity over adjacent intervals, reversal of limits, and the zero-width rule to find unknown integrals from known ones — without ever touching an antiderivative.

Predict: as you drag the split point c left and right, does \(\int_0^c f\,dx + \int_c^8 f\,dx\) change? Drag the dashed split line (or use the slider) and watch the total.

The blue region is \(\int_0^c f\,dx\) and the green region is \(\int_c^8 f\,dx\) — each a net signed area, so parts below the axis count negative. Slide c all the way to 0 to see the zero-width case, and press Reverse the limits to sweep the interval backwards and watch every sign flip.

∫03.0 f dx = 1.875 + ∫3.08 f dx = 2.125 · sum = 4.000 · ∫08 f dx = 4.000

One curve, two pieces — the split moves, the pieces trade value, the total never budges. At c = 0 the blue piece is a zero-width interval: exactly 0.
∫ from 0 to c ∫ from c to 8 split point c (drag me)

The definite integral obeys linearity, additivity over adjacent intervals, reversal of limits (which flips the sign), and equals zero over a zero-width interval — four rules that let you rearrange integrals like ordinary sums.

Formal

For integrable \(f\) and \(g\) and any constant \(k\):
• Linearity: \(\int_a^b [f \pm g]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx\), and \(\int_a^b k f\,dx = k \int_a^b f\,dx\) — split sums, pull out constants, exactly as with derivatives.
• Additivity: \(\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx\) — chop the interval at any shared endpoint and add the pieces. That's what the diagram above shows: the blue piece plus the green piece always equals the whole, no matter where the cut lands.
• Reversal: \(\int_a^b f\,dx = -\int_b^a f\,dx\) — swapping the limits flips the sign, because sweeping backwards makes every \(\Delta x\) negative.
• Zero width: \(\int_a^a f\,dx = 0\) — no width, no area. Each rule falls straight out of the Riemann-sum definition: sums split term by term, factor out constants, concatenate end to end, and vanish when there are no terms.

Worked example

Given \(\int_0^3 f\,dx = 10\) and \(\int_0^5 f\,dx = 4\), find \(\int_3^5 f\,dx\) and \(\int_5^0 f\,dx\).
1. Additivity with the shared endpoint 3: \(\int_0^5 f = \int_0^3 f + \int_3^5 f\), so \(4 = 10 + \int_3^5 f\), giving \(\int_3^5 f\,dx =\) \(-6\). (Yes, negative — the piece from 3 to 5 must pull the running total down from 10 to 4.)
2. Reversal: \(\int_5^0 f\,dx = -\int_0^5 f\,dx =\) \(-4\).

Your turn

Given \(\int_1^4 f\,dx = 7\) and \(\int_4^6 f\,dx = -2\), find \(\int_1^6 f\,dx\) and \(\int_6^1 f\,dx\).
1. The intervals \([1,4]\) and \([4,6]\) share the endpoint 4, so additivity applies: \(\int_1^6 f\,dx = \int_1^4 f\,dx + \int_4^6 f\,dx = 7 + (-2) = \) ____
2. Then reversal gives \(\int_6^1 f\,dx = \) ____

Reveal the answer

\(\int_1^6 f\,dx = 7 + (-2) = \) 5, and \(\int_6^1 f\,dx = -\int_1^6 f\,dx = \) \(-5\). Notice additivity worked even though the second piece was negative — the pieces are net signed areas, and they add as plain numbers.

More info — why reversing the limits flips the sign

Go back to the Riemann sum from the previous two lessons: to integrate from \(a\) to \(b\) you slice the interval into \(n\) steps of width \(\Delta x = (b-a)/n\). If you instead sweep from \(b\) down to \(a\), the step becomes \((a-b)/n = -\Delta x\) — the same heights \(f(x_i^*)\), but every slice now carries a negative width. The whole sum, and so its limit, changes sign: \(\int_b^a f\,dx = -\int_a^b f\,dx\). The zero-width rule is the same idea taken to the extreme: with \(a = b\), \(\Delta x = 0\) and every term of the sum is 0. Try it live — press Reverse the limits in the demo above, or see section 5.2 of the OpenStax text in Dive deeper below.

Check your understanding

Question 1 of 4

If \(\int_2^7 f(x)\,dx = 12\) and \(\int_2^5 f(x)\,dx = 9\), what is \(\int_5^7 f(x)\,dx\)?

Question 2 of 4

Which statement is ALWAYS true for integrable functions \(f\) and \(g\) on \([a,b]\)?

Question 3 of 4

Suppose \(\int_3^8 f(x)\,dx = -5\) — the net signed area is negative because more region sits below the x-axis than above it. What is \(\int_8^3 f(x)\,dx\)?

Question 4 of 4

Given \(\int_0^2 f(x)\,dx = 3\) and \(\int_0^2 g(x)\,dx = -1\), evaluate \(\int_0^2 [2f(x) - 3g(x)]\,dx\).

Recap

  • Linearity: \(\int_a^b [f \pm g]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx\) and \(\int_a^b k f\,dx = k\int_a^b f\,dx\) — split sums, pull out constants. There is no product rule for integrals.
  • Additivity: \(\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx\) — chop at a shared endpoint and add; where you chop doesn't change the total.
  • Reversal: \(\int_a^b f\,dx = -\int_b^a f\,dx\) — sweeping backwards makes every \(\Delta x\) negative, flipping the sign.
  • Zero width: \(\int_a^a f\,dx = 0\) — no width, no area.

Dive deeper

Sources

  • Properties of the Definite Integral