Definite integration for JEE: when symmetry helps—and when it fails

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What you’ll understand

Check the interval and integrability before using reflection, oddness or evenness.

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Quick answer

For an integrable function on $[a,b]$ , reflection gives $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ . Add the two expressions when their sum simplifies. On $[-a,a]$ , an integrable odd function has integral zero and an even function gives twice the integral on $[0,a]$ . Check the domain and singularities before using either shortcut.

Work through the reasoning

Example 1: a reflected pair. Let $I=\int_0^1 \frac{x}{1+x}\,dx$ . Reflection gives $I=\int_0^1 \frac{1-x}{2-x}\,dx$ , but their sum is not constant; reflection alone does not guarantee a short answer. Direct division gives $x/(1+x)=1-1/(1+x)$ , so $I=1-\ln 2$ . Choosing another method after checking is part of good reasoning.

Example 2: a pair that does simplify. For $J=\int_0^1\frac{x^2}{x^2+(1-x)^2}\,dx$ , reflection replaces the numerator by $(1-x)^2$ . Adding gives $2J=\int_0^1 1\,dx=1$ , hence $J=1/2$ . The denominator is positive throughout the interval, so the integrand is continuous.

On zero to one, f(x)=x²/[x²+(1−x)²] rises from zero to one; its reflection f(1−x) falls from one to zero. They sum to one at every x and meet at x=one half. The two integrals are equal and sum to one. — Here f(x) = x²/[x² + (1 − x)²]. The denominator is positive on [0,1]. Reflection leaves the limits unchanged, so each curve has integral ½. The graph illustrates this example, not a rule that every reflected pair sums to a constant.

Common mistakes to catch

Wrong: every odd-looking formula integrates to zero. Instead: verify both the symmetric limits and integrability.
Wrong: $\int_{-1}^{1}1/x\,dx=0$ as an ordinary improper integral. Instead: the separate one-sided integrals diverge. A Cauchy principal value is a different operation.
Wrong: keep applying reflection when it does not simplify the pair. Instead: inspect algebra, substitution or an antiderivative.

Try it yourself

1. Reveal the answer: integrate x³ + x² from −1 to 1.

The odd $x^3$ part contributes zero; the even part contributes $2\int_0^1 x^2\,dx=2/3$ . The polynomial is continuous, so the symmetry rules apply.

2. Reveal the answer: is the ordinary improper integral of 1/x from −2 to 2 zero?

No. Splitting at zero gives divergent one-sided integrals. Symmetry does not make the ordinary improper integral converge. A symmetric principal-value limit is not the same definition.

JEE · Mathematics

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The NCERT source supports the underlying concepts. All examples, self-checks and their explanations here are original editorial exercises, not past-paper questions, frequency estimates or promises of marks. Technical review: Codex (automated); human subject review is not claimed.

NCERT Mathematics: Integrals
NCERT online textbook · Sections on definite integrals and their properties; integrability condition and original counterexample · Checked 2026-10-23

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What you’ll understand

Quick answer

Work through the reasoning

A reflected pair with integral one half

Common mistakes to catch

Try it yourself

Continue studying Definite Integration

Sources and method

Frequently asked questions

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