Question

Let \( f : [1, \infty) \to [2, \infty) \) be a differentiable function. If

\( 10 \int_{1}^{x} f(t) \, dt = 5x f(x) - x^5 - 9 \) for all \( x \ge 1 \), then the value of \( f(3) \) is ______.

• A. 18
• B. 32
• C. 22
• D. 20

Hint:

For this type of problem, use the properties of definite integrals and apply the fundamental theorem of calculus along with the chain rule. Simplify the equation step by step for easy solution.

Answer 1

The Correct Option is B

Given a differentiable function \( f(x) \) defined by an integral equation, determine its value at \( x = 3 \).

Concepts Utilized:

The solution employs the following fundamental concepts:

1. Leibniz Rule (Fundamental Theorem of Calculus, Part 1): This rule facilitates the differentiation of integrals with a variable upper limit. \[ \frac{d}{dx} \int_{a}^{x} g(t) \, dt = g(x) \]
2. Solution of First-Order Linear Differential Equations: Equations in the form \( \frac{dy}{dx} + P(x)y = Q(x) \) are solvable using an integrating factor (I.F.).
   • The Integrating Factor is computed as \( \text{I.F.} = e^{\int P(x) \, dx} \).
   • The general solution is expressed as \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \), where C is the constant of integration.

The approach involves transforming the given integral equation into a differential equation through differentiation, subsequently solving this differential equation to obtain \( f(x) \), and finally calculating \( f(3) \).

Step-by-Step Derivation:

Step 1: Differentiate the provided integral equation with respect to \( x \).

The initial equation is:

\[ 10 \int_{1}^{x} f(t) \, dt = 5x f(x) - x^5 - 9 \]

Applying differentiation to both sides, using the Leibniz rule on the left and the product rule on the \( 5x f(x) \) term on the right:

\[ \frac{d}{dx} \left( 10 \int_{1}^{x} f(t) \, dt \right) = \frac{d}{dx} \left( 5x f(x) - x^5 - 9 \right) \] \[ 10 f(x) = \left( 5 \cdot f(x) + 5x \cdot f'(x) \right) - 5x^4 - 0 \]

Step 2: Restructure the derived equation into a standard first-order linear differential equation.

\[ 10 f(x) = 5f(x) + 5x f'(x) - 5x^4 \]

Simplifying the equation yields:

\[ 5 f(x) = 5x f'(x) - 5x^4 \]

Dividing the entire equation by 5 gives:

\[ f(x) = x f'(x) - x^4 \]

Rearranging into the form \( f'(x) + P(x)f(x) = Q(x) \):

\[ x f'(x) - f(x) = x^4 \]

Given that \( x \geq 1 \), we can divide by \( x \):

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

Step 3: Compute the integrating factor (I.F.) for this linear differential equation.

With \( P(x) = -\frac{1}{x} \), the integrating factor is:

\[ \text{I.F.} = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln|x|} \]

Since \( x \geq 1 \), \( |x| = x \). Thus,

\[ \text{I.F.} = e^{-\ln x} = e^{\ln(x^{-1})} = x^{-1} = \frac{1}{x} \]

Step 4: Determine the general solution of the differential equation.

The solution follows the formula \( f(x) \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \). With \( Q(x) = x^3 \):

\[ f(x) \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} \, dx + C \] \[ \frac{f(x)}{x} = \int x^2 \, dx + C \] \[ \frac{f(x)}{x} = \frac{x^3}{3} + C \]

The general solution for \( f(x) \) is:

\[ f(x) = \frac{x^4}{3} + Cx \]

Step 5: Establish the value of the integration constant C.

The original integral equation is utilized by substituting \( x=1 \) to establish an initial condition.

\[ 10 \int_{1}^{1} f(t) \, dt = 5(1) f(1) - (1)^5 - 9 \]

As the integral from 1 to 1 is zero:

\[ 0 = 5f(1) - 1 - 9 \] \[ 0 = 5f(1) - 10 \implies 5f(1) = 10 \implies f(1) = 2 \]

Substituting \( x=1 \) and \( f(1)=2 \) into the general solution:

\[ 2 = \frac{(1)^4}{3} + C(1) \] \[ 2 = \frac{1}{3} + C \] \[ C = 2 - \frac{1}{3} = \frac{5}{3} \]

Step 6: Formulate the specific solution for \( f(x) \).

Substituting \( C = 5/3 \) back into the general solution yields the particular function:

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

Final Calculation & Result

The objective is to compute \( f(3) \). Substitute \( x=3 \) into the derived function:

\[ f(3) = \frac{(3)^4}{3} + \frac{5}{3}(3) \] \[ f(3) = \frac{81}{3} + 5 \] \[ f(3) = 27 + 5 = 32 \]

Therefore, the value of \( f(3) \) is 32.