Let f(x) be a function satisfying f(x) + f(π - x) = π², ∀x ∈ R. Then integral from 0 to π of f(x) si

Q: Wait, why can't I just find $f(x)$ first? The problem doesn't give me the function!

That is the beauty of this problem! You don't need to know the specific identity of $f(x)$. In JEE Advanced, we often encounter problems where the individual components are unknown, but their collective behavior is defined. By using the King's Property, we bypass the need to solve for $f(x)$ entirely, focusing instead on the symmetry of the integral.

Q: Why does the King's Property work? Is it just magic?

It is not magic, but geometry! Think of the integral as the area under a curve. When you substitute $x$ with $a+b-x$, you are essentially reversing the direction of the x-axis within that interval. The area remains invariant because the shape of the curve is simply being viewed from the other side. It is a fundamental property of definite integrals.

Q: What if the limits were different? Would this still work?

The King's Property is universal for any definite integral $\int_a^b g(x) dx$. However, it is most powerful when the integrand has a symmetry that complements the limits, just like $\sin(\pi-x) = \sin x$ here. Always check if your integrand simplifies after the transformation! If the limits were different, say $0$ to $2\pi$, you would replace $x$ with $2\pi - x$.

Question

Let $f(x)$ be a function satisfying $f(x) + f(\pi - x) = \pi^2, \forall x \in \mathbb{R}$. Then $\int_0^\pi f(x) \sin x \, dx$ is equal to

Options:

(A) $\frac{\pi^2}{4}$

(B) $2\pi^2$

(C) $\pi^2$

(D) $\frac{\pi^2}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

First, we define our integral as:

$$I = \int_{0}^{\pi} f(x) \sin x \, dx \quad \dots (1)$$

We are tasked with finding the value of $I$ without knowing the explicit form of $f(x)$. However, we are provided with the functional constraint:

$$f(x) + f(\pi - x) = \pi^2$$

This symmetry is the key to unlocking the solution.

### The King's Property: The Secret Weapon

The **King's Property** states that for any integrable function $g(x)$:

$$\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$$

Applying this to our integral where $a=0$ and $b=\pi$, we replace $x$ with $(\pi - x)$:

$$I = \int_{0}^{\pi} f(\pi - x) \sin(\pi - x) \, dx$$

Using the trigonometric identity $\sin(\pi - x) = \sin x$, the integral simplifies to:

$$I = \int_{0}^{\pi} f(\pi - x) \sin x \, dx \quad \dots (2)$$

### The Symphony of Cancellation

Now, we add equation (1) and equation (2) together:

$$I + I = \int_{0}^{\pi} f(x) \sin x \, dx + \int_{0}^{\pi} f(\pi - x) \sin x \, dx$$

Combining the integrands under a single integral sign, we obtain:

$$2I = \int_{0}^{\pi} [f(x) + f(\pi - x)] \sin x \, dx$$

Substituting the given constraint $f(x) + f(\pi - x) = \pi^2$ into the expression, the unknown function $f(x)$ is eliminated:

$$2I = \int_{0}^{\pi} \pi^2 \sin x \, dx$$

### The Final Victory

Since $\pi^2$ is a constant, we factor it out of the integral:

$$2I = \pi^2 \int_{0}^{\pi} \sin x \, dx$$

Evaluating the definite integral of $\sin x$:

$$\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$$

Substituting this back into our equation for $2I$:

$$2I = \pi^2 	imes 2$$

Dividing both sides by $2$, we arrive at the final result:

**$I = \pi^2$**

Animated Solution for Mathematics - Definite Integration: Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

Select Answer:

Visualized Solution

Defining the Integral $I$

Applying King's Property

Transforming the Integral

Simplifying $sin (π - x)$

Adding Equations (1) and (2)

Using the Given Identity

Factoring out the Constant

Integrating $sin x$

Evaluating the Limits

Final Calculation

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The King's Property

The Symphony of Cancellation

The Final Victory

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying f(x)+f(π−x)=π2,∀x∈R. Then ∫0π​f(x)sinxdx is equal to

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Visualized Solution

Defining the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I