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Properties of Definite Integrals

Properties of definite integrals transform the limits or integrand to simplify evaluation without first finding a complicated antiderivative.

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Student-friendly explanation

These properties are valid under integrability conditions on the given interval. They are especially powerful for symmetry, interval splitting, reversing limits, and replacing x by a-x on [0,a]. They must be applied with the correct limits and sign.

How to write this in exams

  1. 1

    Start with the exact idea

    Properties of definite integrals transform the limits or integrand to simplify evaluation without first finding a complicated antiderivative.

  2. 2

    Then show how to use it

    Observe the limits first. If limits are reversed, adjust the sign. If the interval is split, use additivity. For 0 to a, try x replaced by a-x. For -a to a, test f(-x). Apply the property only after confirming integrability and simplifying correctly.

  3. 3

    Add one concrete example

    ∫_0^a f(x) dx=∫_0^a f(a-x) dx, provided f is integrable on [0,a].

  4. 4

    Avoid this incomplete answer

    For an odd function on [-a,a], giving 2∫_0^a f(x)dx instead of 0 by confusing odd and even symmetry rules.

Definition

Properties of definite integrals transform the limits or integrand to simplify evaluation without first finding a complicated antiderivative.

Example

∫_0^a f(x) dx=∫_0^a f(a-x) dx, provided f is integrable on [0,a].

Rule to remember

Important properties: ∫_a^b f(x)dx=-∫_b^a f(x)dx; ∫_a^b f(x)dx=∫_a^c f(x)dx+∫_c^b f(x)dx; ∫_0^a f(x)dx=∫_0^a f(a-x)dx; ∫_-a^a f(x)dx=0 if f is odd; ∫_-a^a f(x)dx=2∫_0^a f(x)dx if f is even. Conditions: f must be integrable on the relevant interval.

Memory hook

Limits give the strategy: reversed means sign change, symmetric means check odd or even, 0 to a means try a minus x.

Examples and method

Worked example

Evaluate ∫_0^π x sinx dx using the property I=∫_0^a f(x)dx=∫_0^a f(a-x)dx. Let I=∫_0^π x sinx dx. Also I=∫_0^π (π-x)sin(π-x) dx=∫_0^π (π-x)sinx dx. Adding, 2I=∫_0^π π sinx dx=π[-cosx]_0^π=2π. Hence I=π.

Method to apply

Observe the limits first. If limits are reversed, adjust the sign. If the interval is split, use additivity. For 0 to a, try x replaced by a-x. For -a to a, test f(-x). Apply the property only after confirming integrability and simplifying correctly.

Diagram support

A graph can help with even-odd symmetry. If drawn, label the y-axis as the symmetry line for even functions and show opposite signed areas for odd functions.

How CBSE asks it

Often appears in higher-mark definite integral questions where direct integration is lengthy. The examiner may expect symmetry, interval transformation, or even-odd reasoning.

Avoid common mistakes

Common confusion

Students often reverse limits without changing the sign, writing ∫_a^b f(x)dx=∫_b^a f(x)dx, which is false unless the integral is zero.

Common wrong answer

For an odd function on [-a,a], giving 2∫_0^a f(x)dx instead of 0 by confusing odd and even symmetry rules.

Exam tip

For limits 0 to a, try replacing x by a-x. For limits -a to a, check whether the function is even or odd.

Quick check

What is ∫_-2^2 x^3 dx?

0, because x^3 is odd and the interval is symmetric about 0.

Answer writing and exam use

1-mark answer

Properties of definite integrals transform the limits or integrand to simplify evaluation without first finding a complicated antiderivative.

2-mark answer

Properties of definite integrals transform the limits or integrand to simplify evaluation without first finding a complicated antiderivative. Important properties: ∫_a^b f(x)dx=-∫_b^a f(x)dx; ∫_a^b f(x)dx=∫_a^c f(x)dx+∫_c^b f(x)dx; ∫_0^a f(x)dx=∫_0^a f(a-x)dx; ∫_-a^a f(x)dx=0 if f is odd; ∫_-a^a f(x)dx=2∫_0^a f(x)dx if f is even. Conditions: f must be integrable on the relevant interval. ∫_0^a f(x) dx=∫_0^a f(a-x) dx, provided f is integrable on [0,a].

3-mark answer

These properties are valid under integrability conditions on the given interval. They are especially powerful for symmetry, interval splitting, reversing limits, and replacing x by a-x on [0,a]. They must be applied with the correct limits and sign. Important properties: ∫_a^b f(x)dx=-∫_b^a f(x)dx; ∫_a^b f(x)dx=∫_a^c f(x)dx+∫_c^b f(x)dx; ∫_0^a f(x)dx=∫_0^a f(a-x)dx; ∫_-a^a f(x)dx=0 if f is odd; ∫_-a^a f(x)dx=2∫_0^a f(x)dx if f is even. Conditions: f must be integrable on the relevant interval. Evaluate ∫_0^π x sinx dx using the property I=∫_0^a f(x)dx=∫_0^a f(a-x)dx. Let I=∫_0^π x sinx dx. Also I=∫_0^π (π-x)sin(π-x) dx=∫_0^π (π-x)sinx dx. Adding, 2I=∫_0^π π sinx dx=π[-cosx]_0^π=2π. Hence I=π. Often appears in higher-mark definite integral questions where direct integration is lengthy. The examiner may expect symmetry, interval transformation, or even-odd reasoning. For an odd function on [-a,a], giving 2∫_0^a f(x)dx instead of 0 by confusing odd and even symmetry rules.
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