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 ______.
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 ______.
Show Hint
18
32
22
20
The Correct Option is B
Approach Solution - 1
We are given a differentiable function \( f(x) \) defined by an integral equation, and we need to find the value of this function at \( x = 3 \).
Concept Used:
To solve this problem, we will use the following key concepts:
- Leibniz Rule (Fundamental Theorem of Calculus, Part 1): This rule is used to differentiate an integral with a variable upper limit. \[ \frac{d}{dx} \int_{a}^{x} g(t) \, dt = g(x) \]
- Solving First-Order Linear Differential Equations: An equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) can be solved using an integrating factor (I.F.).
- The Integrating Factor is given by \( \text{I.F.} = e^{\int P(x) \, dx} \).
- The solution is given by \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \), where C is the constant of integration.
The strategy is to first convert the given integral equation into a differential equation by differentiating it, then solve the differential equation to find the function \( f(x) \), and finally calculate \( f(3) \).
Step-by-Step Solution:
Step 1: Differentiate the given integral equation with respect to \( x \).
The given equation is:
\[ 10 \int_{1}^{x} f(t) \, dt = 5x f(x) - x^5 - 9 \]Differentiating both sides with respect to \( x \) using the Leibniz rule for the left side and the product rule for the term \( 5x f(x) \) on the right side:
\[ \frac{d}{dx} \left( 10 \int_{1}^{x} f(t) \, dt \right) = \frac{d}{dx} \left( 5x f(x) - x^5 - 9 \right) \] \[ 10 f(x) = \left( 5 \cdot f(x) + 5x \cdot f'(x) \right) - 5x^4 - 0 \]Step 2: Rearrange the resulting equation to form a first-order linear differential equation.
\[ 10 f(x) = 5f(x) + 5x f'(x) - 5x^4 \]Simplifying the equation:
\[ 5 f(x) = 5x f'(x) - 5x^4 \]Dividing the entire equation by 5:
\[ f(x) = x f'(x) - x^4 \]Rearranging it into the standard form \( f'(x) + P(x)f(x) = Q(x) \):
\[ x f'(x) - f(x) = x^4 \]Since \( x \geq 1 \), we can divide by \( x \):
\[ f'(x) - \frac{1}{x} f(x) = x^3 \]Step 3: Calculate the integrating factor (I.F.) for this linear differential equation.
Here, \( P(x) = -\frac{1}{x} \). The integrating factor is:
\[ \text{I.F.} = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln|x|} \]Since \( x \geq 1 \), \( |x| = x \). So,
\[ \text{I.F.} = e^{-\ln x} = e^{\ln(x^{-1})} = x^{-1} = \frac{1}{x} \]Step 4: Find the general solution of the differential equation.
The solution is given by \( f(x) \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \). 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: Determine the value of the integration constant C.
We use the original integral equation and substitute \( x=1 \) to find an initial condition.
\[ 10 \int_{1}^{1} f(t) \, dt = 5(1) f(1) - (1)^5 - 9 \]Since the integral from 1 to 1 is zero:
\[ 0 = 5f(1) - 1 - 9 \] \[ 0 = 5f(1) - 10 \implies 5f(1) = 10 \implies f(1) = 2 \]Now, substitute \( x=1 \) and \( f(1)=2 \) into our general solution:
\[ 2 = \frac{(1)^4}{3} + C(1) \] \[ 2 = \frac{1}{3} + C \] \[ C = 2 - \frac{1}{3} = \frac{5}{3} \]Step 6: Write the particular solution for \( f(x) \).
Substituting \( C = 5/3 \) back into the general solution, we get the specific function:
\[ f(x) = \frac{x^4}{3} + \frac{5}{3}x \]Final Computation & Result
We are asked to find the value of \( f(3) \). We substitute \( x=3 \) into the function we found:
\[ f(3) = \frac{(3)^4}{3} + \frac{5}{3}(3) \] \[ f(3) = \frac{81}{3} + 5 \] \[ f(3) = 27 + 5 = 32 \]Thus, the value of \( f(3) \) is 32.
Approach Solution -2
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