Properties of Definite Integrals

Compute this integral:

Try finding an antiderivative. The integrand \frac{\sin x}{\sin x + \cos x} does not have a clean closed-form antiderivative in terms of elementary functions — at least not one you would find by standard methods in a reasonable time.

But the answer is \pi/4. And you can get it in three lines, without finding an antiderivative at all, using a single property of definite integrals.

The properties in this article are not just "rules to memorise." They are shortcuts that turn impossible-looking integrals into exercises. Knowing them is the difference between spending 20 minutes on a problem and solving it in 30 seconds.

The basic properties

Here are the foundational rules, each with a proof.

Property 1: Reversing limits

Proof. Let F be an antiderivative of f. Then \int_a^b f(x)\,dx = F(b) - F(a) and \int_b^a f(x)\,dx = F(a) - F(b). These are negatives of each other.

This is the "direction matters" property. Integrating from left to right gives the opposite sign of integrating from right to left. As a special case, \int_a^a f(x)\,dx = 0 — the integral over a zero-width interval is zero.

Property 2: Constant multiple

Proof. If F is an antiderivative of f, then cF is an antiderivative of cf. So \int_a^b cf\,dx = cF(b) - cF(a) = c[F(b) - F(a)] = c\int_a^b f\,dx.

Constants pass through the integral sign. This is the linearity property for scaling.

Property 3: Sum rule

Proof. If F' = f and G' = g, then (F+G)' = f + g. So \int_a^b (f+g)\,dx = [F(b)+G(b)] - [F(a)+G(a)] = [F(b)-F(a)] + [G(b)-G(a)].

Properties 2 and 3 together say the definite integral is a linear operator: \int_a^b [\alpha f + \beta g]\,dx = \alpha\int_a^b f\,dx + \beta\int_a^b g\,dx.

Property 4: Interval splitting

for any c (not necessarily between a and b).

Proof. \int_a^c f\,dx + \int_c^b f\,dx = [F(c) - F(a)] + [F(b) - F(c)] = F(b) - F(a) = \int_a^b f\,dx. The F(c) terms cancel.

This lets you break an integral at any convenient point. It is especially useful when f has a different formula on different parts of the interval (piecewise functions).

Property 5: Comparison

If f(x) \geq g(x) for all x in [a, b], then

Proof. Let h(x) = f(x) - g(x) \geq 0. Each Riemann sum for h is a sum of non-negative terms (each h(x_k^*) \cdot \Delta x \geq 0), so the limit is non-negative. Therefore \int_a^b h\,dx \geq 0, which means \int_a^b f\,dx \geq \int_a^b g\,dx.

A special case: if f(x) \geq 0 on [a, b], then \int_a^b f(x)\,dx \geq 0.

The reflection property

This is the single most useful property for competitive exams.

Proof. In the integral \int_0^a f(x)\,dx, substitute u = a - x. Then du = -dx. When x = 0, u = a; when x = a, u = 0. So:

Rename the dummy variable u back to x:

That is the entire proof — one substitution.

What it means geometrically. Replacing x with a - x reflects the interval [0, a] around its midpoint. The point x maps to a - x: the left endpoint 0 goes to a, the right endpoint a goes to 0, and the midpoint a/2 stays fixed. So the property says: the area under f is the same as the area under the "reflected" function f(a-x).

Solving the opening problem

Return to I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx.

Apply the reflection property with a = \pi/2: replace x with \pi/2 - x.

Now add the original and the reflected:

Three lines. No antiderivative needed.

The generalised reflection property

The same idea works on any interval [a, b], not just [0, a]:

Proof. Substitute u = a + b - x. Then du = -dx, and the limits transform as x = a \to u = b, x = b \to u = a. So \int_a^b f(x)\,dx = \int_b^a f(a+b-u)(-du) = \int_a^b f(a+b-u)\,du.

The special case a = 0 recovers the earlier property.

Even and odd functions on symmetric intervals

When the interval is symmetric about the origin — [-a, a] — and the function has a definite parity, the integral simplifies dramatically.

Proof (even case). Split the integral at 0:

In the first integral, substitute u = -x, so du = -dx:

The last step uses f(-u) = f(u) (the even property). So the two pieces are equal, giving 2\int_0^a f(x)\,dx.

Proof (odd case). The same substitution gives \int_{-a}^0 f(x)\,dx = \int_0^a f(-u)\,du = -\int_0^a f(u)\,du, using f(-u) = -f(u). The two pieces cancel, giving 0.

Practical value. If you see \int_{-2}^2 x^3\,dx, you can immediately say the answer is 0 — the function x^3 is odd. No computation needed. Similarly, \int_{-\pi}^{\pi} \cos x\,dx = 2\int_0^{\pi} \cos x\,dx = 2[\sin x]_0^{\pi} = 0. (The cosine function is even, so we double — and the result still happens to be zero because of where \sin vanishes.)

Common confusions

• "The reflection property only works on [0, a]." The general version works on any [a, b]: replace x with a + b - x. The [0, a] version is just the special case a = 0.

• "I need to know if the function is even or odd to use symmetry." You need the function to be exactly even or odd. If it is neither, the even/odd property does not apply directly. But you can always decompose any function into its even and odd parts: f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}.

• "The integral of an even function is always positive." Not necessarily. \cos x is even, but \int_{-\pi}^{\pi} \cos x\,dx = 0 because the positive and negative parts of the cosine cancel. The even/odd property tells you the integral doubles or vanishes — it says nothing about the sign.

• "I can use the reflection property and the even/odd property together on any integral." They apply to different setups. The reflection property works on [0, a] or [a, b]. The even/odd property works on symmetric intervals [-a, a]. Using one may rearrange the integral into a form where the other applies, but they are distinct tools.

• "Reflection is just a substitution, so it cannot give new information." It is a substitution — but combined with adding the original and reflected integrals, it creates cancellations that would not happen otherwise. The substitution alone gives an equal integral; the addition is where the power lies.

Going deeper

Using comparison to estimate integrals

Property 5 (comparison) lets you bound integrals you cannot compute exactly. For example, on [0, 1], x^2 \leq x (because x \leq 1). So:

More usefully: on [0, 1], 1 \leq e^x \leq e (since e^x is increasing and e^0 = 1, e^1 = e). So:

The exact value is e - 1 \approx 1.718, which is indeed between 1 and e \approx 2.718.

This technique is valuable for JEE problems that ask you to estimate or bound an integral rather than compute it exactly.

Proving properties from the Riemann sum

The proofs above used antiderivatives (the Fundamental Theorem). But every property can also be proved directly from the Riemann sum definition — the limit of \sum f(x_k^*)\Delta x.

For example, the linearity property \int_a^b [f+g]\,dx = \int_a^b f\,dx + \int_a^b g\,dx follows from the fact that \sum [f(x_k^*) + g(x_k^*)]\Delta x = \sum f(x_k^*)\Delta x + \sum g(x_k^*)\Delta x, and taking limits of both sides. No antiderivative is needed.

This matters because not every integrable function has an elementary antiderivative. The properties hold for all Riemann-integrable functions, not just those with nice closed-form antiderivatives. The Riemann sum proofs are more general.

Where this leads next

• Properties - Advanced — periodic function integrals, the \int_0^{2a} property, and the King's property.
• Definite Integration Techniques — substitution and by-parts in definite integrals, with limits tracking.
• Fundamental Theorem of Calculus — the theorem that justifies the antiderivative evaluation used in every proof above.
• Area Under Curves — applying these properties to compute areas bounded by curves.
• Even and Odd Functions — a closer look at symmetry properties of functions.