Let f:R→R be a twice differentiable function such that f(2)=1. If F(x)=xf(x) for all x ∈ R, ∫_0^2 x

Q: Why can't we just integrate $F'(x)$ to get $F(x)$? We don't know the function $F(x)$!

That is exactly the point! In JEE Advanced calculus, we often deal with 'functional' problems where we manipulate the properties of the function rather than finding the function itself. We don't need to know the explicit form of $F(x)$. Trust the process of Integration by Parts to extract the information you need from the structure of the integral.

Q: Why does the lower limit $0$ always seem to vanish in these problems?

It is not a coincidence! Look at the boundary terms like $[x F(x)]_{0}^{2}$. When you plug in the lower limit $x=0$, the term becomes $0 \cdot F(0)$. Unless $F(0)$ is undefined or infinite, this term will always be zero. It is a gift from the problem setter to simplify your life, so embrace it!

Q: I got stuck on the second integral. How do I know when to stop applying Integration by Parts?

Great question. You stop when the resulting integral matches a term you have already solved or when the integral becomes trivial. In this problem, the second IBP step produced a term involving $\int x F'(x) dx$, which we had already calculated. Always look for these 'recursive' connections—they are the bridge to your final answer.

Question

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

Options:

(A) 11

(B) 13

(C) 15

(D) 9

Sigma Percentile · Accepted Answer

### The Calculus of Hidden Patterns

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a detective mission. When you look at a problem like this, it is easy to feel overwhelmed by the notation.

We have a function $f(x)$, a secondary function $F(x) = x f(x)$, and two mysterious integrals involving derivatives. But take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance.

### Phase 1: The Detective's Intuition

We are given $\int_{0}^{2} x F'(x) dx = 6$ and $\int_{0}^{2} x^2 F''(x) dx = 40$. Our goal is to find $F'(2) + \int_{0}^{2} F(x) dx$.

Notice the pattern? We have an algebraic term ($x$ or $x^2$) multiplied by a derivative ($F'(x)$ or $F''(x)$). This is the universal signal for **Integration by Parts (IBP)**.

The IBP formula, $\int u dv = uv - \int v du$, is our most powerful tool for 'peeling' away the layers of a derivative. By choosing $u$ to be the algebraic part, we reduce its power upon differentiation, making the integral simpler.

### Phase 2: The First Breakthrough

Let us tackle the first integral: $\int_{0}^{2} x F'(x) dx = 6$. We set $u = x$ and $dv = F'(x) dx$. This implies $du = dx$ and $v = F(x)$.

Applying the IBP formula, we get:

$$[x F(x)]_{0}^{2} - \int_{0}^{2} F(x) dx = 6$$

Now, evaluate the boundary term $[x F(x)]_{0}^{2}$. At the upper limit, we have $2 F(2)$. At the lower limit, we have $0 \cdot F(0)$, which vanishes into nothingness.

We are left with $2 F(2) - \int_{0}^{2} F(x) dx = 6$. The problem states $F(x) = x f(x)$ and $f(2) = 1$, so $F(2) = 2 \cdot f(2) = 2$.

Substituting this back, we get $2(2) - \int_{0}^{2} F(x) dx = 6$. Solving for the integral, we find:

$$\int_{0}^{2} F(x) dx = 4 - 6 = -2$$

We have successfully captured the first piece of our puzzle!

### Phase 3: The Second Connection

Now, let us look at the second integral: $\int_{0}^{2} x^2 F''(x) dx = 40$. We apply IBP again. Let $u = x^2$ and $dv = F''(x) dx$. Then $du = 2x dx$ and $v = F'(x)$.

The formula gives us:

$$[x^2 F'(x)]_{0}^{2} - \int_{0}^{2} (2x) F'(x) dx = 40$$

Evaluating the boundary term at $x=2$ gives $4 F'(2)$, and at $x=0$ it vanishes. So we have $4 F'(2) - 2 \int_{0}^{2} x F'(x) dx = 40$.

Look closely at that integral term. It is exactly the first integral we solved! We know $\int_{0}^{2} x F'(x) dx = 6$.

Substituting this in, we get $4 F'(2) - 2(6) = 40$. This simplifies to $4 F'(2) - 12 = 40$, or $4 F'(2) = 52$. Dividing by 4, we find:

$$F'(2) = 13$$

### The Final Victory

We have arrived at the finish line. We need the sum of $F'(2)$ and $\int_{0}^{2} F(x) dx$.

We found $F'(2) = 13$ and $\int_{0}^{2} F(x) dx = -2$. Adding them together:

$$13 + (-2) = 11$$

The elegance of this problem lies not in complex computation, but in the recognition of structure. You didn't need to know the function; you only needed to know how to manipulate its properties. Keep this mindset, and no problem will ever be too difficult for you. The final answer is **11**.

Select Answer:

Visualized Solution

Problem Setup

Integration by Parts (IBP)

Applying IBP to the First Integral

Evaluating Limits

Finding the Value of $F (2)$

Solving for the Integral of $F (x)$

Analyzing the Second Integral

Evaluating Limits for the Second Integral

Substituting the Known Integral Value

Calculating $F^{'} (2)$

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Calculus of Hidden Patterns

Phase 1

Phase 2

Phase 3

The Final Victory

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