How to Split a Definite Integral into Two Parts

A definite integral over an interval can be divided into two smaller integrals. If $f$ is integrable on $[a,b]$ and $c$ lies between $a$ and $b$, then

$$\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.$$

The point $c$ can be chosen anywhere in the interval. It does not need to be a zero of the function, a midpoint, or a point where the formula changes. Splitting is often useful when an integrand has different formulas on different intervals, changes sign, has a convenient symmetry, or is easier to integrate in pieces.

Why the rule works

The additivity of the definite integral is a basic property of integration. If $f$ is integrable on $[a,b]$, its restrictions to $[a,c]$ and $[c,b]$ are integrable, and

$$\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.$$

When $f$ is continuous, the same property can also be verified using the Fundamental Theorem of Calculus. Suppose $F$ is an antiderivative of $f$, so that $F'(x)=f(x)$. Then

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

The two smaller integrals give

$$\begin{aligned} \int_a^c f(x)\,dx+\int_c^b f(x)\,dx &=[F(c)-F(a)]+[F(b)-F(c)]\\ &=F(b)-F(a)\\ &=\int_a^b f(x)\,dx. \end{aligned}$$

The middle terms cancel. Geometrically, the signed area from $a$ to $b$ is the signed area from $a$ to $c$ together with the signed area from $c$ to $b$. Areas below the $x$-axis count negatively, so this statement concerns signed area rather than only ordinary positive area.

More generally, if

$$a=x_0<x_1<\cdots<x_n=b,$$

then

$$\int_a^b f(x)\,dx=\sum_{k=1}^{n}\int_{x_{k-1}}^{x_k} f(x)\,dx.$$

Basic worked example

Split and evaluate

$$\int_0^4 (x^2+1)\,dx$$

at $x=1$.

Using additivity,

$$\int_0^4 (x^2+1)\,dx =\int_0^1 (x^2+1)\,dx+\int_1^4 (x^2+1)\,dx.$$

An antiderivative is

$$F(x)=\frac{x^3}{3}+x.$$

The first part is

$$\begin{aligned} \int_0^1 (x^2+1)\,dx &=\left[\frac{x^3}{3}+x\right]_0^1\\ &=\frac13+1=\frac43. \end{aligned}$$

The second part is

$$\begin{aligned} \int_1^4 (x^2+1)\,dx &=\left[\frac{x^3}{3}+x\right]_1^4\\ &=\left(\frac{64}{3}+4\right)-\left(\frac13+1\right)\\ &=\frac{76}{3}-\frac43=24. \end{aligned}$$

Therefore,

$$\int_0^4 (x^2+1)\,dx=\frac43+24=\frac{76}{3}.$$

Computing the original integral directly gives the same result:

$$\left[\frac{x^3}{3}+x\right]_0^4=\frac{64}{3}+4=\frac{76}{3}.$$

Splitting a piecewise-defined integral

Splitting is essential when the integrand has different definitions on different intervals. For example, consider

$$f(x)= \begin{cases} x+1,&0\le x<2,\\ 3x-1,&2\le x\le 4. \end{cases}$$

To evaluate its integral over $[0,4]$, split at the point where the formula changes:

$$\int_0^4 f(x)\,dx =\int_0^2 (x+1)\,dx+\int_2^4 (3x-1)\,dx.$$

Now calculate each part:

$$\int_0^2 (x+1)\,dx =\left[\frac{x^2}{2}+x\right]_0^2=2+2=4,$$

and

$$\begin{aligned} \int_2^4 (3x-1)\,dx &=\left[\frac{3x^2}{2}-x\right]_2^4\\ &=(24-4)-(6-2)=16. \end{aligned}$$

Therefore,

$$\int_0^4 f(x)\,dx=4+16=20.$$

The value assigned at the single point $x=2$ does not affect an ordinary definite integral. What matters is using the correct formula on each interval.

Choosing a useful splitting point

Although any interior point works, some choices simplify the calculation:

• Split where the formula for $f$ changes.
• Split at a discontinuity when the resulting integrals are proper and separately defined.
• Split at a point where a substitution becomes natural.
• Split at the midpoint when using symmetry.
• Split at a zero of $f$ when finding ordinary geometric area rather than signed area.

For example, if $f(x)=x-2$ on $[0,4]$, then

$$\int_0^4 (x-2)\,dx =\int_0^2 (x-2)\,dx+\int_2^4 (x-2)\,dx= -2+2=0.$$

The signed areas cancel. However, the total geometric area between the graph and the axis is

$$\int_0^2 |x-2|\,dx+\int_2^4 |x-2|\,dx=2+2=4.$$

Thus, splitting at zeros is especially useful when converting a signed integral into a total-area calculation.

Improper integrals need extra care

If $f$ is unbounded or undefined at the splitting point, the additivity rule must be interpreted through limits. For example, with $c$ inside $[a,b]$ and a singularity at $c$,

$$\int_a^b f(x)\,dx$$

is usually defined as

$$\int_a^c f(x)\,dx+\int_c^b f(x)\,dx,$$

where each term must converge separately. It is not enough for divergent parts to cancel formally.

For instance,

$$\int_{-1}^{1}\frac{1}{x}\,dx$$

does not converge as an improper integral, because the two one-sided integrals diverge. Its symmetric principal value is $0$, but that is a different concept and should not be confused with an ordinary improper integral.

Common mistakes

1. Forgetting the plus sign. Splitting uses addition, not multiplication:

   $$\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.$$

2. Changing the integrand unnecessarily. If the same formula applies throughout the interval, copy it unchanged into both integrals.

3. Confusing signed area with total area. A negative contribution below the axis remains negative unless you use an absolute value or reverse the sign on that subinterval.

4. Ignoring reversed limits. The convention

   $$\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$$

   still applies, so check the order of every pair of limits.

5. Using cancellation for divergent integrals. For an improper integral, test convergence on each side of a singularity separately.