Splitting integrals is one of the most useful techniques in calculus, and it comes down to a handful of properties you can apply in different situations. Whether you’re breaking a sum into separate integrals, dividing an interval at a specific point, or handling a piecewise function, the core idea is the same: take a complicated integral and turn it into simpler pieces you already know how to solve.
Splitting Sums and Differences
The most common way to split an integral is by separating terms that are added or subtracted. If your integrand is the sum or difference of two functions, you can integrate each one on its own:
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
∫ [f(x) − g(x)] dx = ∫ f(x) dx − ∫ g(x) dx
So an integral like ∫ (x² + sin x) dx becomes ∫ x² dx + ∫ sin x dx, and you can handle each piece independently. This works for any number of terms. An expression with five terms added together can be broken into five separate integrals.
A related property lets you pull constant factors out of the integral entirely:
∫ k·f(x) dx = k · ∫ f(x) dx
The key word here is “constant.” The factor k cannot depend on x. You can pull a 3 or a π out front, but you cannot pull out anything that’s a function of x. Together, these two rules are called the linearity properties of integration, and they’re the workhorses of most integral calculations.
Why You Can’t Split Products or Quotients
A very common mistake is trying to apply the same splitting logic to multiplication or division. The integral of f(x)·g(x) is not equal to the product of their separate integrals. There is no shortcut for ∫ f(x)·g(x) dx the way there is for sums.
For example, ∫ x·e^(6x) dx can’t be broken into ∫ x dx times ∫ e^(6x) dx. That simply gives the wrong answer. Instead, products like this require their own techniques. Integration by parts is the standard tool, using the formula:
∫ u dv = uv − ∫ v du
Other product-type integrands might call for substitution or trigonometric identities. But splitting? Never valid for products or quotients.
Splitting the Interval of Integration
For definite integrals, there’s a second type of splitting that’s completely different from breaking up terms. You can split the interval itself at any point c between a and b:
∫ from a to b of f(x) dx = ∫ from a to c of f(x) dx + ∫ from c to b of f(x) dx
This works because areas are additive. The total area under the curve from a to b is just the area from a to c plus the area from c to b. The function needs to be integrable on the full interval, but c can be any point between a and b. This property might seem abstract until you see the situations where it becomes essential.
Piecewise Functions
Interval splitting is the standard approach for integrating piecewise functions. When a function is defined by different formulas on different parts of the interval, you split the integral at each breakpoint where the formula changes, then use the correct formula on each piece.
Suppose f(x) equals 3x² when x ≤ 1 and equals 6 when x > 1, and you need ∫ from −2 to 3 of f(x) dx. The function changes its definition at x = 1, which falls inside your integration limits. Split there:
∫ from −2 to 3 of f(x) dx = ∫ from −2 to 1 of 3x² dx + ∫ from 1 to 3 of 6 dx
Now each integral contains a single, continuous formula. The first evaluates to x³ from −2 to 1, giving 1 − (−8) = 9. The second evaluates to 6x from 1 to 3, giving 18 − 6 = 12. The total is 21. The recipe is always the same: identify the breakpoints inside your limits, split the integral at each one, and substitute the correct formula for each subinterval.
Absolute Value Functions
Absolute value integrals are really just a special case of piecewise integration, but they trip people up because the breakpoints aren’t handed to you. You have to find them yourself.
There’s no single antiderivative for an absolute value expression. Instead, you use the definition: |expression| equals the expression itself where it’s non-negative, and the negative of the expression where it’s negative. Your job is to figure out where the sign changes.
Consider ∫ from 0 to 4 of |x³ − 5x² + 6x| dx. First, find where the inside equals zero by factoring: x(x − 2)(x − 3) = 0, so the roots are x = 0, 2, and 3. Test each subinterval to determine the sign. On (0, 2) the expression is positive, on (2, 3) it’s negative, and on (3, 4) it’s positive again. Now split:
∫ from 0 to 2 of (x³ − 5x² + 6x) dx + ∫ from 2 to 3 of −(x³ − 5x² + 6x) dx + ∫ from 3 to 4 of (x³ − 5x² + 6x) dx
Notice the negative sign on the middle integral. That’s the absolute value doing its job, flipping the sign where the original expression is negative. Each piece is now a straightforward polynomial integral.
Improper Integrals With Interior Discontinuities
When your integrand has a discontinuity (like a vertical asymptote) somewhere inside the interval of integration, you must split the integral at that point. This isn’t optional. Ignoring an interior discontinuity and just computing the antiderivative across it will give you a wrong answer.
For example, ∫ from 0 to 3 of 1/(x − 1) dx has a discontinuity at x = 1. You split it into two improper integrals:
∫ from 0 to 1 of 1/(x − 1) dx + ∫ from 1 to 3 of 1/(x − 1) dx
Each piece is then evaluated as a limit, with the variable approaching 1 from the left in the first integral and from the right in the second. There’s an important rule here: if either piece diverges (goes to infinity), the entire integral diverges. You cannot cancel a positive infinity from one piece against a negative infinity from the other. That kind of cancellation is not mathematically valid.
Using Symmetry to Simplify
When your interval is symmetric around zero (like −a to a), the symmetry of the function can save you significant work. Even functions satisfy f(x) = f(−x), meaning they’re mirror images across the y-axis. Examples include x², cos x, and |x|. For an even function:
∫ from −a to a of f(x) dx = 2 · ∫ from 0 to a of f(x) dx
Odd functions satisfy f(−x) = −f(x), meaning they have rotational symmetry around the origin. Examples include x³, sin x, and x·cos x. For an odd function integrated over a symmetric interval, the two halves perfectly cancel:
∫ from −a to a of f(x) dx = 0
This result is exact, not an approximation. The negative area on one side cancels the positive area on the other. You can use this to split a complicated integrand into its even and odd parts, immediately dropping any odd terms when the interval is symmetric. The product of an even function and an odd function is always odd, so its integral over a symmetric interval is always zero.
Choosing Where to Split
In most cases, the split point chooses itself. For piecewise functions, you split where the formula changes. For absolute values, you split where the expression inside equals zero. For improper integrals, you split at the discontinuity. For symmetry, you split at zero.
When you have a choice, pick a split point that makes each resulting integral as simple as possible. If part of the interval lets you use a simpler antiderivative or a known result, that’s a natural place to divide. The interval additivity property places no restrictions on where c falls between a and b, so any choice is technically valid. The practical question is just which choice leads to the easiest computation.