Properties of definite integrals

Some definite integrals look impossible to evaluate directly — the antiderivative simply doesn't exist in elementary form — yet they fall apart in two lines once you notice a symmetry. That is the whole point of the properties of definite integrals: a short list of algebraic rules about swapping limits, splitting an interval, and reflecting the variable of integration, all of which follow from nothing more than what "signed area under a curve" means. ISC Class 12 expects you to know each property by name, recognise instantly which one a given integral is begging for, and use the single most powerful trick among them — adding an integral to its own reflection — to crush problems that resist direct integration entirely. This article restates the standard properties, proves the reflection rule geometrically, and drills the "add $I$ to itself" technique on the classic $\sqrt{\sin x}$ example.

The core insight

A definite integral's value depends only on the region under the curve, not on how you label or traverse the interval, so relabelling the variable (reflection, splitting, parity) can turn an unintegrable expression into a trivial one.

The core idea

Let $f$ be continuous on the interval in question. Nothing here needs a new formula for area — every property below is a restatement of "the signed area doesn't care how you describe the interval."

Dummy variable and orientation. The name of the integration variable is irrelevant, and reversing the limits flips the sign:

\int_a^b f(x)\,dx = \int_a^b f(t)\,dt, \qquad \int_a^b f(x)\,dx = -\int_b^a f(x)\,dx

where: $a, b$ are the limits and $x, t$ are dummy labels for the same running variable — swapping the label changes nothing because the variable never appears in the answer.

Splitting the interval. For any point $c$ strictly between $a$ and $b$ :

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx \qquad (a<c<b)

This is the property to reach for whenever the integrand has a "corner" — a modulus, a greatest-integer function, or a piecewise definition — inside the interval.

Reflection about the midpoint. Replacing $x$ by $a+b-x$ inside $f$ leaves the integral unchanged:

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx, \qquad \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

where: the second form is just the special case $a=0$ (renaming the upper limit). This reflection property is the workhorse of the chapter — see the derivation below.

Symmetric intervals and even/odd parity.

\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\displaystyle\int_0^a f(x)\,dx & \text{if } f(-x)=f(x)\ \text{(even)} \\ 0 & \text{if } f(-x)=-f(x)\ \text{(odd)}\end{cases}

and, closely related, $\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f(2a-x)=f(x)$ , and $0$ if $f(2a-x)=-f(x)$ . The figure below makes the reflection rule visual: the curve traversed from $a$ to $b$ and the same curve traversed from $b$ to $a$ under the relabelling $x \to a+b-x$ enclose exactly the same shaded region.

Figure

A curve over an interval and its mirror image reflected about the interval midpoint, with the same shaded area under both, showing why reflecting the variable does not change a definite integral

The reflection property, and the "add I to itself" trick

Derivation AProving the reflection property and using it to add an integral to its own mirror

Prove that $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ for continuous $f$ , and use it to evaluate $I = \int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$ , an integral with no elementary antiderivative that nonetheless yields to this single substitution plus addition.

Step 1. Substitute to prove the reflection identity. In $\int_a^b f(a+b-x)\,dx$ , let $u = a+b-x$ , so $du = -dx$ . When $x=a$ , $u=b$ ; when $x=b$ , $u=a$ . The integral becomes $\int_b^a f(u)(-du) = \int_a^b f(u)\,du$ , which is the same number as $\int_a^b f(x)\,dx$ since the name of the variable does not matter. This proves the property for any continuous $f$ — no special structure of $f$ is needed.

Step 2. Apply it to $I$ with $a=0$ , $b=\pi/2$ , so the substitution is $x \to \dfrac{\pi}{2}-x$ . Using $\sin\left(\dfrac{\pi}{2}-x\right)=\cos x$ and $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$ :

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

Both expressions equal $I$ — the reflection just relabels the same area.

Step 3. Add the two equal expressions for $I$ . The denominators are identical, so the numerators add directly:

2I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx + \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx = \int_0^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}

Step 4. Solve for $I$ . Since $2I = \pi/2$ , $I = \dfrac{\pi}{4}$ . Notice the original integrand never had to be antidifferentiated at all — the reflection property converted an intractable integral into the integral of the constant $1$ . Result: $I = \pi/4$ , matching the Pass-1 worked example.

Worked example

Evaluate $\displaystyle\int_{-2}^{2} x^3\cos x\,dx$ and $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2x\,dx$ using the even/odd property, extending the earlier parity check to a full symmetric-interval calculation.

Step 1. Test the parity of $g(x) = x^3\cos x$ . Compute $g(-x) = (-x)^3\cos(-x) = -x^3\cos x = -g(x)$ , since $\cos$ is even. So $g$ is odd, and by the symmetric-interval property $\int_{-a}^a f\,dx = 0$ for odd $f$ :

\int_{-2}^2 x^3\cos x\,dx = 0

No antiderivative work was needed — the odd symmetry alone kills the integral.

Step 2. Test the parity of $h(x) = \cos^2x$ . Compute $h(-x) = \cos^2(-x) = \cos^2x = h(x)$ , so $h$ is even, and the property halves the interval while doubling the coefficient:

\int_{-\pi/2}^{\pi/2}\cos^2x\,dx = 2\int_0^{\pi/2}\cos^2x\,dx

Step 3. Now integrate the halved range using $\cos^2x = \dfrac{1+\cos2x}{2}$ :

2\int_0^{\pi/2}\frac{1+\cos2x}{2}\,dx = \left[x+\frac{\sin2x}{2}\right]_0^{\pi/2} = \left(\frac{\pi}{2}+\frac{\sin\pi}{2}\right)-(0+0) = \frac{\pi}{2}

Both answers: $\displaystyle\int_{-2}^2 x^3\cos x\,dx = \mathbf{0}$ and $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2x\,dx = \boldsymbol{\dfrac{\pi}{2}}$ .

Competitive stretch

For $\displaystyle\int_0^2 |x-1|\,dx$ , the corner at $x=1$ forces interval-splitting rather than parity: for $x\in[0,1]$ , $|x-1|=1-x$ ; for $x\in[1,2]$ , $|x-1|=x-1$ . Splitting at $c=1$ gives $\int_0^1(1-x)\,dx + \int_1^2(x-1)\,dx = \frac12 + \frac12 = 1$ . The lesson: pick the property that matches the integrand's structure — parity for symmetric smooth functions, splitting for anything with a corner inside the interval.

Common traps

Using the $\int_0^{2a}$ or $\int_{-a}^{a}$ symmetry shortcuts without actually checking the required condition — $f(2a-x)=f(x)$ , or genuine even/odd parity — instead of testing $f(-x)$ algebraically first.
Substituting $x \to a-x$ or $x \to a+b-x$ but forgetting to replace every occurrence of $x$ inside $f$ , leaving a mixed, wrong expression.
Missing the moment to apply the "add $I$ to its reflection" trick — look for integrands built from $\sqrt{\sin x}, \sqrt{\cos x}$ , or any ratio that becomes its own complement under $x \to a-x$ .
Splitting an interval at a point outside $(a,b)$ , or failing to notice a hidden corner in a modulus or greatest-integer integrand (like $|x-1|$ or $[x]$ ) that silently needs a split even when the problem doesn't flag it.

Ask the tutor: Not sure which property fits your integral, or want to check whether a reflection-and-add trick will actually cancel — walk through the integrand with the AI tutor.

How do I know which property to try first on an unfamiliar integral?

Scan the integrand for a giveaway shape before reaching for anything else. A modulus, greatest-integer function, or a piecewise rule inside the interval means split it at the corner. A symmetric interval like $[-a,a]$ means test $f(-x)$ for even/odd parity. A ratio or sum built from $\sqrt{\sin x}, \sqrt{\cos x}$ , or an expression that looks "complementary" to itself under $x \to a-x$ is a strong hint to try the reflection-and-add trick. If none of these fit, the integral probably just needs a direct antiderivative.

Why does adding I to its own reflection ever simplify anything?

Because the reflection substitution $x \to a+b-x$ does not change the value of the integral, only how the integrand is written — so you get a second, algebraically different-looking expression that is provably equal to the same $I$ . Adding two expressions that are individually complicated but structurally complementary (like $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$ and $\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}$ ) can make the messy parts cancel, leaving something trivial like the constant $1$ . It only works when the sum genuinely simplifies — check that before committing.

Can I apply the even/odd shortcut to any integral over a symmetric interval?

Only if you actually verify the parity first. Compute $f(-x)$ and compare it to $f(x)$ : if they're equal, $f$ is even and the integral is twice the integral over $[0,a]$ ; if $f(-x)=-f(x)$ , $f$ is odd and the integral is zero. Many students apply the "symmetric interval means zero" shortcut without checking, which is wrong whenever the function isn't actually odd — for example $\cos^2x$ on $[-\pi/2,\pi/2]$ is even, not odd, and equals $\pi/2$ , not $0$ .