Problem 185 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(a+b-x)=f(x)\), then \(\int_{a}^{b} x f(x) d x\) is equal to (A) \(\frac{a+b}{2} \int_{a}^{b} f(b-x) d x\) (B) \(\frac{a+b}{2} \int_{a}^{b} f(x) d x\) (C) \(\frac{b-a}{2} \int_{a}^{b} f(x) d x\) (D) \(\frac{a+b}{2} \int_{a}^{b} f(a+b-x) d x\)

Step-by-Step Solution

Answer

Option (B): \(\frac{a+b}{2} \int_{a}^{b} f(x) dx\).

Step 1: Analyze the given condition

We have the condition \( f(a+b-x) = f(x) \). This means that the function \( f \) is symmetric about the line \( x = \frac{a+b}{2} \). This symmetry will play a crucial role in solving the integral.

Step 2: Express x in terms of a and b

Recognize that each \( x \) can be rewritten as \( (a+b) - (a+b-x) \). This will help when transforming the integral, as integrating \( x f(x) \) over \( [a, b] \) with substitution can help simplify it due to the symmetry condition of \( f \).

Step 3: Change of Variable in the Integral

Consider changing variables in the integral. Let \( u = a + b - x \), then \( du = -dx \), which means that as \( x \) goes from \( a \) to \( b \), \( u \) goes from \( b \) to \( a \). This will allow us to utilize the given symmetric property of \( f \) in the integral.

Step 4: Integrate with respect to new variable

Using the change of variables, the integral \( \int_{a}^{b} x f(x) dx = \int_{b}^{a} [(a+b)-u] f((a+b)-u) (-du) \). Simplifying gives \( = \int_{a}^{b} [(a+b)-u] f(u) du \).

Step 5: Simplify the symmetric integral

Notice that \( [(a+b)-u] = a+b-u \). The integral can be written as: \( \int_{a}^{b} [(a+b)-u] f(u) du = (a+b) \int_{a}^{b} f(u) du - \int_{a}^{b} u f(u) du \). Applying the property \( f(a+b-u) = f(u) \) again helps simplify the expression.

Step 6: Final Step: Evaluate the expression

Perform the algebraic manipulations to simplify the result. \( 2 \int_{a}^{b} x f(x) dx = (a+b) \int_{a}^{b} f(x) dx \). So, \( \int_{a}^{b} x f(x) dx = \frac{a+b}{2} \int_{a}^{b} f(x) dx \). This corresponds to option (B).

Key Concepts

Symmetry in Functions

Symmetry in functions is an important concept in calculus, especially in definite integrals. When a function is symmetric about a certain line, it means the function mirrors itself with respect to that line. In this exercise, we have the condition \( f(a+b-x) = f(x) \), which tells us that the function \( f \) is symmetric around the line \( x = \frac{a+b}{2} \).
This symmetry can simplify the evaluation of integrals significantly. Instead of computing the integral directly, we can exploit the symmetry to reduce complexity or find important properties of the integral. In symmetric functions, the area calculated from \( a \) to \( \frac{a+b}{2} \) is mirrored from \( \frac{a+b}{2} \) to \( b \), allowing us to predict and confirm integral behavior geometrically or algebraically, reducing computation obstacles.

Integral Calculus Substitution

Integral calculus substitution involves changing variables to simplify the process of integration. In our case, we use the substitution \( u = a + b - x \), giving us \( du = -dx \). This technique lets us take advantage of the symmetry property provided in the integral.
When \( x \) moves from \( a \) to \( b \), \( u \) correspondingly moves from \( b \) to \( a \).

• We reverse the limits of integration since \( du = -dx \), effectively maintaining the direction of integration.
• Substitution transforms the integral \( \int_{a}^{b} x f(x) dx \) into a new form \( \int_{b}^{a} \left((a+b)-u\right) f((a+b)-u)(-du) \).

This change of variables neatly aligns the integral with the given symmetry property, making it simpler to solve and revealing underlying algebraic relations.

Properties of Definite Integrals

Understanding the properties of definite integrals is key to solving complex integrals. The properties of symmetry and substitution show how integrals can be manipulated for simplicity and accuracy.
Definite integrals provide the net area under a curve, and by applying properties like symmetry and change of variable, we make these calculations more manageable. Here, the definite integral of interest simplifies due to symmetry, allowing us to split and re-combine the integral.
For example, by recognizing \( \int_{a}^{b} [(a+b)-u] f(u) du = (a+b) \int_{a}^{b} f(u) du - \int_{a}^{b} u f(u) du \), we use the symmetry idea \( f(a+b-u) = f(u) \) again, ultimately simplifying the expression to derive \( \int_{a}^{b} x f(x) dx = \frac{a+b}{2} \int_{a}^{b} f(x) dx \). This illustrates how properties streamline problem-solving and provide solutions that are consistent and verifiable.