Key Concepts and Formulas

• Leibniz Integral Rule: For a function $g(x) = \int_{a}^{x} h(t) \, dt$, its derivative is $g'(x) = h(x)$. This is a direct consequence of the Fundamental Theorem of Calculus Part 1.
• Product Rule for Differentiation: If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$. This is essential for differentiating terms involving the product of $x$ and $f(x)$.
• Linear First-Order Differential Equations: An equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ can be solved using an Integrating Factor (I.F.) given by $e^{\int P(x) dx}$. Multiplying the equation by the I.F. results in the derivative of $(y \cdot \text{I.F.})$.

Step-by-Step Solution

Step 1: Differentiate the Given Integral Equation

We are given the equation $10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9$ for all $x \geqslant 1$. To eliminate the integral and obtain a differential equation, we differentiate both sides with respect to $x$.

Differentiating the left-hand side (LHS) using the Leibniz Integral Rule: $\frac{d}{dx} \left( 10 \int_1^x f(t) \, dt \right) = 10 \cdot f(x)$

Differentiating the right-hand side (RHS) requires the product rule for the term $5x f(x)$ and the power rule for $x^5$: $\frac{d}{dx} \left( 5x f(x) - x^5 - 9 \right) = \left( \frac{d}{dx}(5x) \cdot f(x) + 5x \cdot \frac{d}{dx}(f(x)) \right) - \frac{d}{dx}(x^5) - \frac{d}{dx}(9)$ $= (5 \cdot f(x) + 5x \cdot f'(x)) - 5x^4 - 0$ $= 5f(x) + 5x f'(x) - 5x^4$

Equating the derivatives of both sides: $10 f(x) = 5f(x) + 5x f'(x) - 5x^4$

Step 2: Formulate a Linear First-Order Differential Equation

Rearrange the equation obtained in Step 1 to obtain a standard differential equation form. Subtract $5f(x)$ from both sides: $5f(x) = 5x f'(x) - 5x^4$ Divide the entire equation by $5$: $f(x) = x f'(x) - x^4$ Let $y = f(x)$, so $f'(x) = \frac{dy}{dx}$. Substituting these into the equation: $y = x \frac{dy}{dx} - x^4$ Rearrange to the standard linear first-order form $\frac{dy}{dx} + P(x)y = Q(x)$. First, isolate the terms with $y$ and $\frac{dy}{dx}$: $x \frac{dy}{dx} - y = x^4$ Since $x \ge 1$, $x \neq 0$, we can divide by $x$: $\frac{dy}{dx} - \frac{1}{x} y = x^3$ This is a linear first-order differential equation with $P(x) = -\frac{1}{x}$ and $Q(x) = x^3$.

Step 3: Solve the Linear Differential Equation

Calculate the Integrating Factor (I.F.): $\text{I.F.} = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|}$ Since $x \ge 1$, $|x|=x$, so: $\text{I.F.} = e^{-\ln x} = e^{\ln(x^{-1})} = x^{-1} = \frac{1}{x}$ Multiply the differential equation $\frac{dy}{dx} - \frac{1}{x} y = x^3$ by the I.F. $\frac{1}{x}$: $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = x^3 \cdot \frac{1}{x}$ $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = x^2$ The left-hand side is the derivative of the product of $y$ and the I.F.: $\frac{d}{dx} \left( y \cdot \frac{1}{x} \right) = x^2$ Integrate both sides with respect to $x$: $\int \frac{d}{dx} \left( y \cdot \frac{1}{x} \right) dx = \int x^2 dx$ $y \cdot \frac{1}{x} = \frac{x^3}{3} + C$ Where $C$ is the constant of integration. Solve for $y$ by multiplying by $x$: $y = x \left( \frac{x^3}{3} + C \right) = \frac{x^4}{3} + Cx$ Substituting back $y = f(x)$: $f(x) = \frac{x^4}{3} + Cx$

Step 4: Determine the Constant of Integration (C)

To find the value of $C$, we use the original integral equation and substitute $x=1$. $10 \int_1^1 f(t) \, dt = 5(1) f(1) - (1)^5 - 9$ Since $\int_1^1 f(t) \, dt = 0$: $10 \cdot 0 = 5 f(1) - 1 - 9$ $0 = 5 f(1) - 10$ $5 f(1) = 10 \implies f(1) = 2$ Now, substitute $x=1$ and $f(1)=2$ into the general solution $f(x) = \frac{x^4}{3} + Cx$: $2 = \frac{(1)^4}{3} + C(1)$ $2 = \frac{1}{3} + C$ $C = 2 - \frac{1}{3} = \frac{6-1}{3} = \frac{5}{3}$ Thus, the particular solution is: $f(x) = \frac{x^4}{3} + \frac{5}{3}x$

Step 5: Calculate the Value of $f(3)$

Substitute $x=3$ into the particular solution for $f(x)$: $f(3) = \frac{(3)^4}{3} + \frac{5}{3}(3)$ $f(3) = \frac{81}{3} + 5$ $f(3) = 27 + 5$ $f(3) = 32$

Common Mistakes & Tips

• Incorrectly applying the Product Rule: Ensure all parts of the product rule are used when differentiating $5xf(x)$.
• Forgetting the constant of integration: The constant $C$ is crucial and must be determined using an initial condition derived from the original equation.
• Using the differentiated equation to find the initial condition: Always go back to the original integral equation to find conditions like $f(1)$, as substituting into the differentiated equation might miss information.

Summary

The problem involves transforming an integral equation into a first-order linear differential equation by differentiating both sides using the Leibniz Integral Rule and the product rule. The resulting differential equation is solved using an integrating factor. An initial condition, $f(1)=2$, is obtained by substituting $x=1$ into the original integral equation. This condition allows us to find the specific solution for $f(x)$, which is then used to calculate $f(3)$. The value of $f(3)$ is found to be 32.

The final answer is $\boxed{32}$.