Problem 83
Question
. Suppose that \(f(t)\) has the property that \(f^{\prime}(a)=f^{\prime}(b)=0\) and that \(f(t)\) has two continuous derivatives. Use integration by parts to prove that \(\int_{a}^{b} f^{\prime \prime}(t) f(t) d t \leq 0 .\) Hint \(:\) Use integration by parts by differentiating \(f(t)\) and integrating \(f^{\prime \prime}(t) .\) This result has many applications in the field of applied mathematics.
Step-by-Step Solution
Verified Answer
\(\int_{a}^{b} f''(t) f(t) dt \leq 0 \) because \(\int_{a}^{b} f'(t)^2 dt \geq 0\).
1Step 1: Set up the integration by parts formula
The integration by parts formula is given by \( \int u \cdot v' \, dt = uv - \int v \cdot u' \, dt \). Here, choose \( u = f(t) \) and \( v' = f''(t) \). This means \( u' = f'(t) \) and \( v = f'(t) \).
2Step 2: Apply integration by parts
Substitute the expressions for \( u \), \( v \), and their derivatives into the formula: \[ \int_{a}^{b} f(t) f''(t) \, dt = \left[ f(t)f'(t) \right]_{a}^{b} - \int_{a}^{b} f'(t)^2 \, dt. \]
3Step 3: Evaluate the boundary terms
Evaluate \( \left[ f(t)f'(t) \right]_{a}^{b} = f(b)f'(b) - f(a)f'(a) \). Since \( f'(a) = 0 \) and \( f'(b) = 0 \), this expression simplifies to \( 0 - 0 = 0 \).
4Step 4: Simplify the integral
The expression becomes: \[ \int_{a}^{b} f(t) f''(t) \, dt = - \int_{a}^{b} f'(t)^2 \, dt. \]
5Step 5: Analyze the result for negativity
Since \( f'(t)^2 \) is always non-negative, \( -\int_{a}^{b} f'(t)^2 \, dt \leq 0 \) because the integral of a non-negative function is non-negative, thus negating it produces a non-positive result.
Key Concepts
Continuous DerivativesBoundary ConditionsApplied Mathematics
Continuous Derivatives
Continuous derivatives are derivatives that exist and are smooth, meaning they don’t have any abrupt changes, over the entire domain of a function. In simpler terms, if the derivative of a function doesn’t "jump" or "break" at any point, it is continuous. For the given problem, where the function \( f(t) \) has two continuous derivatives, it ensures a smooth behavior for both \( f'(t) \) and \( f''(t) \). This smoothness is crucial when applying techniques like integration by parts. It guarantees that integration can be carried out smoothly without encountering any discontinuities.
**Why is this important?**
**Why is this important?**
- Ensures no undefined behavior.
- Validates the use of calculus operations without issues.
- Enables accurate estimation or modeling of real-world phenomena.
Boundary Conditions
Boundary conditions are specific requirements or constraints placed on the values of a function at the edges of its domain. In the exercise, the boundary conditions are that the derivative \( f'(t) \) equals zero at the points \( a \) and \( b \). This means that at these "boundaries," the rate of change of the function is zero, pointing to critical points such as a peak, valley, or inflection point.
**Significance in Integration by Parts:**
In the proof, these boundary conditions dramatically simplify the integration by parts result by turning the expression \([f(t)f'(t)]_{a}^{b}\) into zero. Thus, boundary conditions can greatly influence the outcomes of function evaluations and transformations.
**Significance in Integration by Parts:**
- They allow the boundary terms to cancel out simplifying calculations.
- Turn the integration limits into zero, which leads to a simpler expression.
In the proof, these boundary conditions dramatically simplify the integration by parts result by turning the expression \([f(t)f'(t)]_{a}^{b}\) into zero. Thus, boundary conditions can greatly influence the outcomes of function evaluations and transformations.
Applied Mathematics
Applied mathematics uses mathematical techniques and principles to solve real-world problems, which is why it’s deeply intertwined with fields like physics, engineering, and economics. In the provided exercise, integration by parts showing \( \int_{a}^{b} f''(t) f(t) dt \leq 0 \) is vital within numerous applied mathematics scenarios. For instance, energy minimization problems, stability analysis, and more.
**Integration by Parts in Applied Settings:**
**Integration by Parts in Applied Settings:**
- Can model physical systems and derive equations for various phenomena.
- Helps verify stability and feasibility within dynamic systems.
- Assists in deducing optimal conditions for system performance.
Other exercises in this chapter
Problem 81
Show that $$ f(t)=f(a)+\sum_{i=1}^{n} \frac{f^{(i)}(a)}{i !}(t-a)^{i}+\int_{a}^{t} \frac{(t-x)^{n}}{n !} f^{(n+1)}(x) d x $$ provided that \(f\) can be differen
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The Beta function, which is important in many branches of mathematics, is defined as $$ B(\alpha, \beta)=\int_{0}^{1} x^{\alpha-1}(1-x)^{\beta-1} d x $$ with th
View solution Problem 84
Derive the formula $$ \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t=\int_{0}^{x} f(t)(x-t) d t $$ using integration by parts.
View solution Problem 86
If \(P_{n}(x)\) is a polynomial of degree \(n\), show that $$ \int e^{x} P_{n}(x) d x=e^{x} \sum_{j=0}^{n}(-1)^{j} \frac{d^{j} P_{n}(x)}{d x^{j}} $$
View solution