Let for some function y=f(x), ∫_0^x t f(t) dt = x² f(x), x 0 and f(2) = 3 Then f(6) is equal to
(1

Q: Wait, why do we use the Product Rule on the right side? Can't we just differentiate $x^2$ and $f(x)$ separately?

That is a common trap! The Product Rule is mandatory because $x^2$ and $f(x)$ are both functions of $x$. If you differentiate them separately, you are essentially treating $f(x)$ as a constant, which it is not. Always remember: if two terms containing the variable of differentiation are multiplied, the Product Rule is your best friend.

Q: What happened to the constant of integration $C$ when we solved for $f(x)$?

Great observation! When we integrated, we got $\ln|f(x)| = -\ln|x| + C$. Using the property $\ln(a) + \ln(b) = \ln(ab)$, this becomes $\ln|x f(x)| = C$. Taking the exponential of both sides gives $|x f(x)| = e^C$. Since $e^C$ is just another constant, we replace it with $k$. Thus, $x f(x) = k$, or $f(x) = k/x$. The constant $C$ is effectively absorbed into $k$.

Q: Why did we assume $x > 0$ in the problem?

The problem explicitly states $x > 0$. This is crucial because the natural logarithm function $\ln(x)$ is only defined for positive values. If $x$ were negative, we would need to handle the absolute value signs more carefully, but here, the constraint simplifies our life significantly!

Question

Let for some function $y=f(x)$, $\int_{0}^{x}tf(t)dt=x^{2}f(x)$, $x>0$ and $f(2)=3$ Then $f(6)$ is equal to

Options:

(A) 1

(B) 3

(C) 6

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a locked vault. Inside, there is a function $f(x)$ that you desperately need to find, hidden behind the integral equation:

$$\int_{0}^{x} t f(t) \, dt = x^{2} f(x)$$

This is a challenge to your intuition. To liberate a function that is both inside an integral and outside, multiplied by $x^2$, we must utilize the power of calculus.

### The Skeleton Key: Leibniz's Rule

To free $f(x)$, we need a tool that bridges the gap between the integral and the derivative: the Leibniz Integral Rule. By differentiating both sides of the equation with respect to $x$, we determine how the area under the curve $t f(t)$ changes as the boundary $x$ moves.

According to the Fundamental Theorem of Calculus, the derivative of the left side is simply the integrand evaluated at the upper limit:

$$\frac{d}{dx} \int_{0}^{x} t f(t) \, dt = x f(x)$$

Now, we address the right side, $x^2 f(x)$, using the Product Rule. The derivative is:

$$\frac{d}{dx} [x^2 f(x)] = f(x) \cdot \frac{d}{dx}(x^2) + x^2 \cdot \frac{d}{dx}(f(x)) = 2x f(x) + x^2 f'(x)$$

### The Dance of Variables

Equating the two sides, we obtain:

$$x f(x) = 2x f(x) + x^2 f'(x)$$

Rearranging the terms to group the $f(x)$ components yields:

$$-x f(x) = x^2 f'(x)$$

Assuming $x 
eq 0$, we divide both sides by $x^2 f(x)$ to arrive at the separable differential equation:

$$\frac{f'(x)}{f(x)} = -\frac{1}{x}$$

### The Final Integration

We integrate both sides with respect to $x$:

$$\int \frac{f'(x)}{f(x)} \, dx = -\int \frac{1}{x} \, dx$$

This yields $\ln|f(x)| = -\ln|x| + C$. Using the properties of logarithms, we rewrite this as $\ln|f(x)| + \ln|x| = C$, which simplifies to $\ln|x f(x)| = C$.

Taking the exponential of both sides, we find $x f(x) = e^C$. Letting $e^C = k$, where $k$ is a constant, we obtain the general family of functions:

$$f(x) = \frac{k}{x}$$

### Solving for the Specific Function

We use the initial condition $f(2) = 3$ to find the specific member of this family. Substituting $x=2$ and $f(2)=3$ into the general form:

$$3 = \frac{k}{2} \implies k = 6$$

Our function is revealed as:

$$f(x) = \frac{6}{x}$$

Finally, calculating $f(6)$ is a simple matter of substitution:

$$f(6) = \frac{6}{6} = 1$$

The final answer is **1**.

Animated Solution for Mathematics - Differential Equations: Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

Select Answer:

Visualized Solution

The Integral Equation

Applying Leibniz's Rule

Differentiating the LHS

Differentiating the RHS

Forming the Differential Equation

Separating Variables

Integrating Both Sides

Solving for $f (x)$

Using the Initial Condition

The Final Function

Calculating $f (6)$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Skeleton Key

The Dance of Variables

The Final Integration

Solving for the Specific Function

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let for some function y=f(x), ∫0x​tf(t)dt=x2f(x), x>0 and f(2)=3 Then f(6) is equal to