Let p(x) be a function defined on R such that p'(x) = p'(1-x), for all x ∈ [0, 1], p(0) = 1 and p(1)

Q: Wait, why do we get a negative sign when integrating $p'(1-x)$? Is it just a rule?

It is the Chain Rule in reverse! When you integrate a function of the form $f(g(x))$, you must account for the derivative of $g(x)$. Here, $g(x) = 1-x$, so $g'(x) = -1$. Dividing by this $-1$ is exactly why that negative sign appears. It is not just a rule; it is the fundamental geometry of scaling the input variable.

Q: Can I just assume $p(x)$ is a linear function like $ax+b$?

While you might get the correct answer here, assuming linearity is a dangerous habit in JEE Advanced. The problem doesn't state $p(x)$ is linear. The beauty of this method is that it works for *any* function satisfying the condition, whether it's a polynomial, a trigonometric function, or something else entirely. Always aim for the general proof!

Q: Why does the King's Property work? It feels like magic.

It is not magic; it is a change of variables! If you substitute $u = 1-x$, the limits of integration flip, and the differential $dx$ becomes $-du$. The negative sign from the differential flips the limits back to $0$ to $1$. It is a beautiful geometric symmetry where you are essentially reading the area under the curve from right to left instead of left to right.

Question

Let $p(x)$ be a function defined on $\mathbb{R}$ such that $p'(x) = p'(1-x)$, for all $x \in [0, 1], p(0) = 1$ and $p(1) = 41$. Then $\int_{0}^{1} p(x) dx$ equals

Options:

(A) 21

(B) 41

(C) 42

(D) $\sqrt{41}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a calculus problem; we are embarking on a journey of discovery. We are looking at a function $p(x)$ defined on the interval $[0, 1]$.

We know where it starts, $p(0) = 1$, and we know where it ends, $p(1) = 41$. The problem provides a cryptic clue: $p'(x) = p'(1-x)$.

This condition is the heartbeat of the problem. It is a statement of symmetry, a mirror image in the world of derivatives. Let us peel back the layers of this mystery together.

### The Derivative Mirror

Imagine you are standing on a graph of $p(x)$. The condition $p'(x) = p'(1-x)$ tells us that the slope of the curve at any point $x$ is exactly the same as the slope at the point $1-x$.

If you are at $x=0.1$, the steepness of the curve is identical to the steepness at $x=0.9$. To understand the function $p(x)$, we must move from the world of slopes back to the world of values through integration.

We integrate both sides of the equation with respect to $x$:

$$\int p'(x) dx = \int p'(1-x) dx$$

On the left, the integral of $p'(x)$ is simply $p(x)$. On the right, we must apply the Chain Rule; since the derivative of the inner function $(1-x)$ is $-1$, we must divide by this factor. This yields:

$$p(x) = -p(1-x) + C$$

By rearranging, we find a beautiful functional equation:

$$p(x) + p(1-x) = C$$

### Finding the Constant

We have a constant $C$ that we do not know yet. However, we have the boundary conditions $p(0) = 1$ and $p(1) = 41$. If we plug $x=0$ into our new equation, we get:

$$p(0) + p(1-0) = C$$

$$p(0) + p(1) = C$$

$$1 + 41 = C \Rightarrow C = 42$$

The sum of the function values at symmetric points $x$ and $1-x$ is always $42$. Our functional equation is $p(x) + p(1-x) = 42$.

### The King's Property

Now, we turn our attention to the goal: finding the area under the curve, $I = \int_{0}^{1} p(x) dx$. We invoke the 'King's Property' of definite integrals:

$$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$$

Applying this to our integral with $a=1$, we see that:

$$I = \int_{0}^{1} p(1-x) dx$$

We now have two expressions for the same area $I$. Adding them together gives:

$$2I = \int_{0}^{1} [p(x) + p(1-x)] dx$$

### The Grand Finale

Look at the term inside the integral. We already proved that $p(x) + p(1-x) = 42$. The complexity vanishes, and the integral becomes:

$$2I = \int_{0}^{1} 42 dx$$

Integrating a constant is straightforward. The integral of $42$ from $0$ to $1$ is $42(1-0) = 42$.

$$2I = 42$$

**$I = 21$**

The area under the curve is **$21$**. We did not need to know the exact form of $p(x)$ or solve a complex differential equation; we simply used the symmetry provided by the derivative and the elegance of the King's Property.

Animated Solution for Mathematics - Definite Integration: Let $p (x)$ be a function defined on $R$ such that $p^{'} (x) = p^{'} (1 - x)$ , for all $x \in [0, 1], p (0) = 1$ and $p (1) = 41$ . Then $\int_{0}^{1} p (x) d x$ equals

Select Answer:

Visualized Solution

The Setup and The Goal

Decoding the Derivative Symmetry

Integrating Both Sides

Applying the Chain Rule

The Functional Equation

Finding the Constant of Integration

Evaluating $C$

Enter King's Property

Applying King's Property

Adding the Integrals

Substituting the Functional Equation

The Final Computation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Derivative Mirror

Finding the Constant

The King's Property

The Grand Finale

Similar Questions

Let $f_{n} = \int_{0}^{π /2} (\sum_{k = 1}^{n} sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) sin^{k - 1} x) cos x d x, n \in N$ . Then $f_{21} - f_{20}$ is equal to

Let $f : R \to R$ be a twice differentiable function such that $f (2) = 1.$ If $F (x) = x f (x)$ for all $x \in R$ , $\int_{0}^{2} x F^{'} (x) d x = 6$ and $\int_{0}^{2} x^{2} F^{''} (x) d x = 40$ then $F^{'} (2) + \int_{0}^{2} F (x) d x$ is equal to:

Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

The value(s) of $\int_{0}^{1} \frac{x ^{4} ( 1 - x ) ^{4}}{1 + x ^{2}} d x$ is (are)

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

Let $α \in (0, 1)$ and $β = lo g_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + ... + \frac{x ^{n}}{n}, x \in (0, 1)$ . Then the integral $\int_{0}^{α} \frac{t ^{50}}{1 - t} d t$ is equal to

Let $f$ be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2$ , for all $x \in R$ . Then $\int_{0}^{3} f (x) d x$ is equal to

The integral $\int_{- 1/2}^{1/2} ([x] + ln (\frac{1 + x}{1 - x})) d x$ equal to

Let $f : [- 1, 2] \to [0, \infty)$ be a continuous function such that $f (x) = f (1 - x)$ for all $x \in [- 1, 2]$ . Let $R_{1} = \int_{- 1}^{2} x f (x) d x$ , and $R_{2}$ be the area of the region bounded by $y = f (x)$ , $x = - 1$ , $x = 2$ , and the x-axis. Then

The value of the integral $\int_{0}^{1} \frac{1 - x}{1 + x} d x$ is