Let $f : [1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_1^x f(t)dt = 3xf(x) + x^3 – 4$ for all $x \geq 1$, then the value of $f(2) – f(3)$ is :
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Solution Steps
1
Find Initial Condition $f(1)$
Substitute $x = 1$ in the given equation $6 \int_1^x f(t)dt = 3xf(x) + x^3 – 4$.
$6 \int_1^1 f(t)dt = 3(1)f(1) + 1^3 – 4$
$0 = 3f(1) – 3 \implies 3f(1) = 3 \implies f(1) = 1$.
2
Differentiate using Leibniz Rule
Differentiating both sides with respect to $x$:
$\frac{d}{dx} \left[ 6 \int_1^x f(t)dt \right] = \frac{d}{dx} \left[ 3xf(x) + x^3 – 4 \right]$
$6f(x) = 3f(x) + 3xf'(x) + 3x^2$
3
Rearrange into Linear Differential Equation
Subtracting $3f(x)$ and dividing by $3x$:
$3f(x) – 3x^2 = 3xf'(x)$
$xf'(x) – f(x) = -x^2 \implies f'(x) – \frac{1}{x}f(x) = -x$
This is of the form $\frac{dy}{dx} + Py = Q$.
4
Calculate Integrating Factor (I.F.)
$P = -\frac{1}{x}$, so $I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.
5
Solve for $f(x)$
$f(x) \cdot \frac{1}{x} = \int (-x) \cdot \frac{1}{x} dx + C$
$\frac{f(x)}{x} = -x + C$. Using $f(1) = 1$:
$1 = -1 + C \implies C = 2$.
So, $f(x) = x(2 – x) = 2x – x^2$.
6
Calculate Final Value
$f(2) = 2(2) – 2^2 = 4 – 4 = 0$
$f(3) = 2(3) – 3^2 = 6 – 9 = -3$
$f(2) – f(3) = 0 – (-3) = 3$.
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Key Insight
Whenever you see an integral from a constant to $x$ in a functional equation, start by differentiating using the Leibniz rule to convert it into a differential equation.
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Calculus Theory
1. Newton-Leibniz Formula
The fundamental theorem of calculus allows us to differentiate an integral where the limits are functions of the independent variable. Specifically, if $F(x) = \int_{a}^{x} f(t) dt$, then $F'(x) = f(x)$. In more complex cases where both limits are functions of $x$, say $g(x)$ and $h(x)$, the derivative is $f(h(x))h'(x) – f(g(x))g'(x)$. This rule is the primary tool for solving integral equations by transforming them into differential equations. It bridges the gap between the accumulation of a function (integration) and its rate of change (differentiation).
2. Linear Differential Equations
A first-order linear differential equation has the standard form $y’ + P(x)y = Q(x)$. These equations are solved using the “Integrating Factor” (I.F.) method. The I.F. is defined as $e^{\int P(x) dx}$. When the entire equation is multiplied by this factor, the left-hand side becomes the exact derivative of the product $y \cdot I.F.$. This reduces the calculus problem to a standard integration task. In this problem, the variable coefficient $-1/x$ led to an I.F. of $1/x$, which simplified the expression $xf'(x) – f(x)$ effectively.
3. Initial Conditions in Integral Equations
Integral equations involving $\int_{a}^{x}$ often lack an explicit constant of integration. However, they contain a hidden “initial condition.” By setting $x$ equal to the lower limit of the integral ($x=a$), the integral term becomes zero because the area under a point is zero. This provides a specific value for the function (e.g., $f(1)=1$), which is necessary to solve for the constant of integration $C$ that arises after solving the differential equation. Without this step, the general solution remains undetermined.
4. Functional and Differentiable Properties
The problem defines $f$ as a differentiable function on $[1, \infty)$. This is a prerequisite for applying the Leibniz rule. Furthermore, the domain $[1, \infty)$ ensures that values like $1/x$ are well-defined and that we are working within a continuous interval. When solving for values like $f(2)$ or $f(3)$, we must ensure that our derived function $f(x)$ is consistent with the original integral equation over the entire domain. The final result is a specific numerical value derived from the properties of the recovered polynomial function.
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FAQs
Q
What if the lower limit was not 1?
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You would substitute that specific value for x to find the initial condition. For example, if it was ∫₀ˣ, you’d use x=0.
Q
Can I solve the integral first?
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No, because $f(t)$ is unknown. You must differentiate to remove the integral sign.
Q
Why did $e^{-\ln x}$ become $1/x$?
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By log properties, $-\ln x = \ln(x^{-1}) = \ln(1/x)$. Since $e$ and $\ln$ are inverses, $e^{\ln(1/x)} = 1/x$.
Q
Is the product rule used in Step 2?
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Yes, $\frac{d}{dx}[3xf(x)]$ requires the product rule: $3[1 \cdot f(x) + x \cdot f'(x)]$.
Q
What is the value of f(3)?
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Plugging $x=3$ into $f(x) = 2x – x^2$ gives $2(3) – 9 = -3$.
Q
Is f(x) always a polynomial in these problems?
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Not necessarily. Depending on the right-hand side, it could be exponential or trigonometric.
Q
Why differentiate both sides?
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To eliminate the integral and create a differential equation which is easier to solve for $f(x)$.
Q
What is the range of f(x) for x ≥ 1?
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For $f(x) = 2x – x^2$, the vertex is at (1, 1). So for $x \geq 1$, $f(x) \in (-\infty, 1]$.
Q
Can f(x) be non-differentiable?
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The problem states $f$ is differentiable, which is consistent with our result $2x-x^2$.
Q
Why is the answer positive 3?
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Since $f(2)=0$ and $f(3)=-3$, the difference $f(2) – f(3) = 0 – (-3) = 3$.
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