Let f : [1, ∞) → [2, ∞) be a differentiable function, If 10 ∫₁ˣ f(t)dt = 5xf(x) - x⁵ - 9 for all x ≥

Q: Why do we need to find $f(1)$ before differentiating?

The differentiation process introduces a constant of integration $C$. Without an initial condition, we would have an infinite family of solutions. Finding $f(1)$ pins down the specific function we need.

Q: Is it always $1/x$ for the integrating factor?

Not always! The integrating factor depends entirely on the coefficient of $f(x)$ in your differential equation. Here, it happened to be $-1/x$, leading to $1/x$. Always derive it carefully.

Q: Can I use substitution instead of the integrating factor?

In this specific case, the equation $f'(x) - \frac{1}{x}f(x) = x^3$ is a standard linear form. While you might try other methods, the Integrating Factor method is the most robust and elegant way to handle this structure.

Question

Let $f: [1, \infty) 	o [2, \infty)$ be a differentiable function, If $10 \int_1^x f(t)dt = 5xf(x) - x^5 - 9$ for all $x \ge 1$, then the value of $f(3)$ is :

Options:

(A) 18

(B) 32

(C) 22

(D) 26

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the integral equation:

$$10 \int_1^x f(t)dt = 5xf(x) - x^5 - 9$$

This equation defines the relationship between the function $f(x)$ and its integral. Our goal is to determine the specific form of $f(x)$ and evaluate it at a given point.

### Phase 1: The Initial Condition

To find our starting point, we set $x=1$. The integral term vanishes because the limits of integration become identical:

$$10 \int_1^1 f(t)dt = 5(1)f(1) - 1^5 - 9$$

Since the integral from $1$ to $1$ is $0$, the equation simplifies to $0 = 5f(1) - 10$. This immediately reveals our anchor point:

**$f(1) = 2$**

### Phase 2: The Calculus Transformation

To solve for $f(x)$, we differentiate both sides of the original equation with respect to $x$ using the Leibniz Rule. The left side becomes $10f(x)$.

Applying the product rule to the right side, specifically to the term $5xf(x)$, we obtain:

$$10f(x) = 5(f(x) + xf'(x)) - 5x^4$$

Expanding and simplifying this expression, we arrive at:

$$10f(x) = 5f(x) + 5xf'(x) - 5x^4$$

### Phase 3: The ODE Journey

Rearranging the terms to isolate the derivative, we obtain a first-order linear differential equation:

$$5f(x) = 5xf'(x) - 5x^4$$

Dividing by $5x$ (assuming $x 
eq 0$), we get:

$$f'(x) - \frac{1}{x}f(x) = x^3$$

We now determine the Integrating Factor ($I.F.$):

$$I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$$

Multiplying the ODE by the $I.F.$ transforms the equation into a total derivative:

$$\frac{d}{dx} \left( \frac{f(x)}{x} ight) = x^2$$

### Phase 4: The Final Victory

Integrating both sides with respect to $x$ yields:

$$\frac{f(x)}{x} = \frac{x^3}{3} + C$$

Using our initial condition $f(1) = 2$, we substitute $x=1$ to find $C$:

$$\frac{2}{1} = \frac{1}{3} + C \Rightarrow C = \frac{5}{3}$$

The specific function is therefore:

$$f(x) = \frac{x^4}{3} + \frac{5x}{3}$$

Finally, evaluating the function at $x=3$:

$$f(3) = \frac{3^4}{3} + \frac{5(3)}{3} = \frac{81}{3} + 5 = 27 + 5$$

**$f(3) = 32$**

Animated Solution for Mathematics - Differential Equations: Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function, If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of $f (3)$ is :

Select Answer:

Visualized Solution

The Integral Equation

Finding $f (1)$

Differentiating via Leibniz Rule

Differentiating the RHS

Equating Derivatives

Forming the Differential Equation

Identifying P and Q

Finding the General Solution

Solving for Constant $C$

Calculating $f (3)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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