Properties of definite integrals

A definite integral like $\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$ looks impossible to antidifferentiate directly — there is no elementary formula for that integrand. Yet it has a clean closed-form value, and finding it needs no antiderivative at all. The trick is a small toolkit of properties of definite integrals: rules about swapping limits, splitting an interval, and reflecting $x$ about the midpoint of $[a,b]$ , all of which leave the value of the integral unchanged while changing its appearance. Recognising which property turns an ugly integrand into something that cancels, or into an integrand you already know, is the real skill of this chapter — it is a distinct habit of mind from the routine "find the antiderivative, plug in the limits" approach of ordinary evaluation. This article collects the standard properties, proves the workhorse one, and drills the signature technique of adding an integral to its own transformed version.

The core insight

The value of $\int_a^b f(x)\,dx$ is unchanged under reversing limits, splitting at an interior point, or replacing $x$ by $a+b-x$ inside $f$ — and combining the last of these with the original integral, then adding, often collapses a hard integrand to something trivial.

The core idea

Definite integrals are shape-invariant under a few legal rewrites. Let $f,g$ be integrable on the relevant intervals and $a,b,c \in \mathbb{R}$ . The standard list used in this chapter:

\int_a^b f(x)\,dx = \int_a^b f(t)\,dt \qquad \text{(P1: the integration variable is a dummy)}

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx \qquad \text{(P2: reversing limits flips the sign)}

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx, \quad a<c<b \qquad \text{(P3: split at an interior point)}

where: in P3, $c$ is any point strictly between $a$ and $b$ — in practice, the point where a piecewise formula (like $|x|$ ) changes its rule.

Two properties do the heaviest lifting because they let you re-express an integral in a different-looking but equal form, and adding the two forms often kills the awkward part of the integrand:

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \qquad \text{(P4: reflect about the midpoint of }[a,b]\text{)}

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \qquad \text{(P5: the }a=0\text{ special case of P4)}

Symmetric intervals get a shortcut of their own, exploiting whether $f$ is even or odd:

\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\displaystyle\int_0^a f(x)\,dx, & f(-x)=f(x)\ \text{(even)} \\[4pt] 0, & f(-x)=-f(x)\ \text{(odd)} \end{cases} \qquad \text{(P6)}

where: the interval must be exactly $[-a,a]$ , symmetric about $0$ — P6 does not apply to, say, $[-1,3]$ . Testing even/odd means computing $f(-x)$ algebraically, not guessing from how the formula looks. The figure below shows why P6 works geometrically: an even function's graph is a mirror image across the $y$ -axis, so the left half and right half contribute equal area; an odd function's graph is a point-reflection through the origin, so the left half is the exact negative of the right half and the two areas cancel.

Figure

Two graphs on symmetric intervals from negative a to a, an even function whose mirrored left and right areas add, and an odd function whose left and right areas are equal magnitude but opposite sign and cancel to zero

Derivation

Derivation AReflecting the integrand about the midpoint of the interval

The goal is to prove property P4, $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ , purely by substitution, since this single property (with $a=0$ giving P5) is what makes the "add the integral to its own transform" technique legal, and every hard-looking integral in this chapter is cracked using exactly this reflection.

Step 1. Start from the right-hand side and substitute $u = a+b-x$ . Then $du = -dx$ , so $dx = -du$ . When $x=a$ , $u = a+b-a = b$ ; when $x=b$ , $u = a+b-b = a$ . The right-hand side becomes $\int_a^b f(a+b-x)\,dx = \int_{u=b}^{u=a} f(u)\,(-du)$ .

Step 2. Flip the limits to absorb the minus sign, using P2. Reversing the limits of integration changes the sign of the integral, so $\int_{u=b}^{u=a} f(u)\,(-du) = \int_{u=a}^{u=b} f(u)\,du$ .

Step 3. Rename the dummy variable. By P1, the name of the integration variable does not affect the value, so $\int_a^b f(u)\,du = \int_a^b f(x)\,dx$ , which is exactly the left-hand side we started with.

Step 4. Conclude. Chaining Steps 1 to 3 gives $\int_a^b f(a+b-x)\,dx = \int_a^b f(x)\,dx$ , which is P4. Setting $a=0$ gives the frequently-used special case P5, $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ . Both are exact identities for any integrable $f$ — no symmetry of $f$ itself is required, unlike P6.

Worked example

Evaluate $\displaystyle I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$ .

Step 1. Apply P5 with $a=\pi/2$ , using $\sin\!\left(\frac{\pi}{2}-x\right)=\cos x$ and $\cos\!\left(\frac{\pi}{2}-x\right)=\sin x$ :

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

Step 2. Add this to the original expression for $I$ . The two fractions share a denominator, and their numerators add up to it exactly:

2I = \int_0^{\pi/2} \left(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right)dx = \int_0^{\pi/2} 1\,dx.

Step 3. Evaluate the trivial remaining integral and solve for $I$ :

2I = \big[x\big]_0^{\pi/2} = \frac{\pi}{2} \quad\Rightarrow\quad I = \frac{\pi}{4}.

Answer: $I = \dfrac{\pi}{4}$ .

Competitive stretch

The same reflect-and-add idea handles integrands with a bare $x$ in front, using the extra fact $a+b-x$ replaces $x$ too. Evaluate $J = \int_0^{\pi} \dfrac{x}{1+\sin x}\,dx$ . Apply P5 with $a=\pi$ : $J = \int_0^\pi \dfrac{\pi-x}{1+\sin(\pi-x)}\,dx = \int_0^\pi \dfrac{\pi-x}{1+\sin x}\,dx$ , since $\sin(\pi-x)=\sin x$ . Adding this to the original $J$ cancels the two $x$ -dependent numerators into a constant $\pi$ : $2J = \pi\int_0^\pi \dfrac{dx}{1+\sin x}$ . That remaining integral is a standard one (multiply numerator and denominator by $1-\sin x$ , use $1-\sin^2x=\cos^2x$ ) evaluating to $2$ , so $2J = 2\pi$ and $J=\pi$ . The pattern — replace $x$ by $a+b-x$ , add, watch the awkward part vanish — is identical to the main example; only the bookkeeping is longer.

Common traps

Using the even/odd shortcut P6 on an interval that is not exactly $[-a,a]$ — it never applies to something like $[-1,3]$ .
Misjudging even/odd by appearance instead of computing $f(-x)$ algebraically (e.g. $x^2\sin x$ is odd, since $(-x)^2\sin(-x) = -x^2\sin x$ ).
Dropping the sign when reversing limits (P2), or when splitting an interval that effectively runs backwards.
Stopping after applying P4/P5 once instead of adding the transformed integral to the original and solving — using P5 alone rarely finishes a problem; the technique only works because the sum simplifies.

Ask the tutor: Want more practice spotting which property applies, or a walkthrough of the reflect-and-add trick on a trickier integral? Ask the AI tutor for extra worked examples on properties of definite integrals.

Why does replacing x by a+b-x not change the value of the integral?

Because it is really just a substitution in disguise. Setting $u=a+b-x$ flips the direction of integration (high $x$ becomes low $u$ and vice versa), but reversing the limits of integration flips the sign back, and renaming the dummy variable from $u$ back to $x$ changes nothing else. The three effects exactly cancel, so the integral's value is provably unchanged — this is what the derivation above shows step by step.

How do I know whether to use P4/P5 or the even/odd rule P6 on a given integral?

Look at the limits first. If they are symmetric about zero, of the form $-a$ to $a$ , test whether the integrand is even or odd — P6 often finishes the problem in one line. If the limits are $0$ to $a$ (or any $a$ to $b$ ) and the integrand looks hard to antidifferentiate directly, especially if it contains square roots of sine and cosine or similar symmetric-looking pieces, try P5 or P4 and see if adding the original and reflected forms cancels the awkward part.

Do I need to memorise the substitution proof of P4, or just the formula?

For CBSE exams, you mainly need to apply P4 and P5 correctly, not reproduce the substitution proof from scratch every time — but understanding it is what stops you from misapplying the property (for instance, forgetting that both the integrand and the limits transform together). If a question asks you to "prove" a property, the substitution $u=a+b-x$ followed by P2 and P1, exactly as in the derivation above, is the expected method.