Problem 14
Question
Given that \(h(a)=h(b)\). Then the value of \(\int_{a}^{b}[f\\{g[h(x)]\\}]^{-1} f^{\prime}\\{g[h(x)]\\} g^{\prime}[h(x)] h^{\prime}(x) d x\) is (A) 0 (B) \(\frac{h(a)+h(b)}{2}\) (C) \(g(h(b))-g(h(a))\) (D) \(\log [f\\{g[h(b)]\\}]-\log [f\\{g[h(a)]\\}]\)
Step-by-Step Solution
Verified Answer
The value of the integral is 0.
1Step 1: Understand the problem
We are given an integral expression and need to evaluate it given the condition that \( h(a) = h(b) \). This condition implies that the values of the function \( h(x) \) at \( a \) and \( b \) are equal.
2Step 2: Analyze the integral
The integral expression is \( \int_{a}^{b} [f\{g[h(x)]\}]^{-1} f^{\prime}\{g[h(x)]\} g^{\prime}[h(x)] h^{\prime}(x) dx \). This looks like an expression derived from the chain rule and potentially related to the derivative of a composition of functions.
3Step 3: Use substitution for simplification
Let's set \( u = g[h(x)] \). Then \( \frac{du}{dx} = g^{\prime}[h(x)] h^{\prime}(x) \). This transforms the integral into \( \int [f(u)]^{-1} f^{\prime}(u) du \).
4Step 4: Evaluate the simplified integral
The integral \( \int [f(u)]^{-1} f^{\prime}(u) du \) can be simplified to \( \int \, d(\log |f(u)|) \) based on the derivative of the natural logarithm. The result would be \( \log | f(u) | \bigg|_{u=g(h(a))}^{u=g(h(b))} \).
5Step 5: Consider h(a) = h(b) condition
Since \( h(a) = h(b) \), \( g(h(a)) = g(h(b)) \) as well. This implies that at both points, the function values are equal, leading to: \( \log | f(g(h(b))) | - \log | f(g(h(a))) | = 0 \).
6Step 6: Conclusion
Thus, the evaluation of the given integral under the condition \( h(a) = h(b) \) results in 0.
Key Concepts
Chain RuleFunction CompositionSubstitution MethodLogarithmic Integration
Chain Rule
The Chain Rule is a fundamental principle in calculus essential for differentiating composite functions.
It allows us to understand how a change in one variable affects another within a composition of functions. In simple terms, if you have a function composed of two or more functions, the Chain Rule helps find its derivative.Consider a composite function like \( f(g(x)) \). The Chain Rule tells us that the derivative of this function is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). Mathematically, it can be expressed as:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]This principle is vital for simplifying complex expressions, especially when integrating or differentiating functions nested within each other. Understanding the Chain Rule enables a smoother transition between differentiation and integration.
It allows us to understand how a change in one variable affects another within a composition of functions. In simple terms, if you have a function composed of two or more functions, the Chain Rule helps find its derivative.Consider a composite function like \( f(g(x)) \). The Chain Rule tells us that the derivative of this function is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). Mathematically, it can be expressed as:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]This principle is vital for simplifying complex expressions, especially when integrating or differentiating functions nested within each other. Understanding the Chain Rule enables a smoother transition between differentiation and integration.
Function Composition
Function Composition involves creating a new function by applying one function to the results of applying another function.
This is a crucial concept in calculus, as many real-world problems require working with layers of functions.When you compose two functions, say \( g \) and \( h \), you end up with \( g(h(x)) \) as a new function. Function composition allows us to model complex relationships and dynamic systems.
This is a crucial concept in calculus, as many real-world problems require working with layers of functions.When you compose two functions, say \( g \) and \( h \), you end up with \( g(h(x)) \) as a new function. Function composition allows us to model complex relationships and dynamic systems.
- First, apply the inner function \( h \) to an input \( x \).
- Then, take the result of \( h(x) \) and use it as the input for the function \( g \).
Substitution Method
The Substitution Method
is a strategy in integral calculus used to simplify complex integrals, making them more manageable to solve. This approach is particularly handy when directly finding an integral is not straightforward.This method involves selecting a part of the integral to substitute with a new variable, reducing it to a simpler form. In the context of the original exercise:
is a strategy in integral calculus used to simplify complex integrals, making them more manageable to solve. This approach is particularly handy when directly finding an integral is not straightforward.This method involves selecting a part of the integral to substitute with a new variable, reducing it to a simpler form. In the context of the original exercise:
- We let \( u = g[h(x)] \).
- Then, the differential \( du \) aligns with \( g'(h(x)) \cdot h'(x) \), allowing a seamless transition.
Logarithmic Integration
Logarithmic Integration is a technique used in calculus to find integrals involving inverse functions or products.
This involves leveraging the properties of logarithms to simplify the integration process.In the context of our exercise, after substitution, the integral simplifies to an expression involving the derivative of a logarithmic function: \[ \int \frac{f'(u)}{f(u)} \, du = \int \, d(\log |f(u)|) \]The integration becomes a straightforward computation of the natural logarithm function.Logarithmic Integration is potent because it decomposes the integration process into simpler terms. Using the properties of logarithms, it breaks down the integral of complicated expressions into more manageable parts. This technique is beneficial when dealing with exponential growth/decay processes or when encountering natural logarithmic functions within integrals.
This involves leveraging the properties of logarithms to simplify the integration process.In the context of our exercise, after substitution, the integral simplifies to an expression involving the derivative of a logarithmic function: \[ \int \frac{f'(u)}{f(u)} \, du = \int \, d(\log |f(u)|) \]The integration becomes a straightforward computation of the natural logarithm function.Logarithmic Integration is potent because it decomposes the integration process into simpler terms. Using the properties of logarithms, it breaks down the integral of complicated expressions into more manageable parts. This technique is beneficial when dealing with exponential growth/decay processes or when encountering natural logarithmic functions within integrals.
Other exercises in this chapter
Problem 12
The value of the integral \(\int_{0}^{41 \pi / 4}|\cos x| d x\) is (A) \(20-\frac{1}{\sqrt{2}}\) (B) \(20+\frac{1}{\sqrt{2}}\) (C) \(19+\frac{1}{\sqrt{2}}\) (D)
View solution Problem 13
If \(\int_{0}^{y} e^{-t^{2}} d t+\int_{0}^{x^{2}} \sin ^{2} t d t=0\), then \(\frac{d y}{d x}\) is equal to (A) \(2 x \sin ^{2} x^{2} e^{y^{2}}\) (B) \(-2 x \si
View solution Problem 15
\(\int_{0}^{\infty} x^{n} e^{-x} d x(n\) is \(\mathrm{a}+v e\) integer \()\) is equal to (A) \(n !\) (B) \((n-1) !\) (C) \((n-2) !\) (D) None of these
View solution Problem 16
The area of the smaller part of the circle \(x^{2}+y^{2}=a^{2}\), cut off by the line \(x=\frac{a}{\sqrt{2}}\), is given by (A) \(\frac{a^{2}}{2}\left(\frac{\pi
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