Let f be a polynomial function such that f(x² + 1) = x⁴ + 5x² + 2, for all x ∈ R. Then ∫(0 to 3) f(x

Q: Wait, why can't I just integrate $x^4 + 5x^2 + 2$ from $0$ to $3$ directly?

That is a classic trap! The expression $x^4 + 5x^2 + 2$ is equal to $f(x^2+1)$, not $f(x)$. Integrating the composite function directly would give you the wrong area because the input variable is transformed. You must first isolate $f(x)$ before performing the integration.

Q: When I substitute $u = x^2+1$, do I need to change the limits of integration from $0$ to $3$?

Great question! You only change the limits if you are integrating with respect to $u$ (i.e., $\int f(u) du$). Here, we are using substitution to find the *form* of the function $f(x)$. Once we have $f(x) = x^2 + 3x - 2$, we simply integrate that function with respect to $x$ using the original limits $0$ to $3$.

Q: Is it always valid to just replace $u$ with $x$ at the end?

Yes, absolutely. In the context of a polynomial function, $u$ is just a dummy variable representing the input. Whether you call it $f(u)$, $f(x)$, or $f(\text{anything})$, the functional rule remains the same. Once you have the expression in terms of $u$, replacing it with $x$ gives you the standard definition of the function.

Question

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

Options:

(A) $\frac{5}{3}$

(B) $\frac{41}{3}$

(C) $\frac{27}{2}$

(D) $\frac{33}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we encounter a classic problem that tests not just your calculus skills, but your ability to see through the 'mask' of a function.

We are given a functional equation: $f(x^2 + 1) = x^4 + 5x^2 + 2$.

At first glance, it looks like a simple polynomial, but the input $x^2 + 1$ is a transformation. To find the integral $\int_{0}^{3} f(x) dx$, we must first strip away this transformation and reveal the true identity of $f(x)$.

### The Substitution Strategy

We cannot integrate $f(x^2+1)$ directly because the variable of integration is $x$, not $x^2+1$. We need to simplify the input.

Let us introduce a new variable, $u$, such that $u = x^2 + 1$. This is our key to the kingdom.

If $u = x^2 + 1$, then it follows that $x^2 = u - 1$. Furthermore, $x^4$ is simply $(x^2)^2$, which becomes $(u - 1)^2$. By making these substitutions, we transform the entire equation into a function of $u$.

### The Algebraic Transformation

Now, let us substitute these into our original equation:

$$f(u) = (u - 1)^2 + 5(u - 1) + 2$$

This is where precision matters. Expanding $(u - 1)^2$ gives us $u^2 - 2u + 1$. Distributing the $5$ gives us $5u - 5$.

Adding the constant $2$, we get the full expression:

$$f(u) = (u^2 - 2u + 1) + (5u - 5) + 2$$

Combining like terms, we find that $-2u + 5u = 3u$, and $1 - 5 + 2 = -2$. Thus, $f(u) = u^2 + 3u - 2$.

Since $u$ is just a placeholder, we can write **$f(x) = x^2 + 3x - 2$**. We have successfully unmasked the function!

### The Final Integration

With $f(x) = x^2 + 3x - 2$ in hand, the integral $\int_{0}^{3} (x^2 + 3x - 2) dx$ becomes a straightforward exercise in the power rule.

We integrate term by term:

$$\int_{0}^{3} (x^2 + 3x - 2) dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_{0}^{3}$$

Evaluating this from $0$ to $3$, we plug in the upper limit:

$$\left( \frac{3^3}{3} + \frac{3(3^2)}{2} - 2(3) \right) = \left( 9 + \frac{27}{2} - 6 \right) = 3 + 13.5 = 16.5$$

Converting this to a fraction, we get the final result: **$\frac{33}{2}$**.

### A Moment of Reflection

Look at the elegance of this result. We started with a complex-looking composite function and, through the power of substitution and careful algebra, reduced it to a simple quadratic.

This is the essence of JEE Advanced mathematics: taking a seemingly impossible problem and breaking it down into manageable, logical steps. You have mastered the functional transformation. Keep this clarity of thought, and no problem will ever be too daunting. Onward!

Animated Solution for Mathematics - Definite Integration: 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

Select Answer:

Visualized Solution

Analyze the Functional Equation

Substitution for Simplification

Expressing $f (u)$ in terms of $u$

Expanding the Polynomial

Simplifying to find $f (x)$

Setting up the Integral

Applying Integration Rules

Evaluating at Upper Limit

Evaluating at Lower Limit

Final Arithmetic and Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Substitution Strategy

The Algebraic Transformation

The Final Integration

A Moment of Reflection

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