Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Answer

The value of the integral is 0, corresponding to option (A).

Step 1: Understand the given integral

The integral given in the problem is \( \int_{a}^{b}[f\{g[h(x)]\}]^{-1} f^{\prime}\{g[h(x)]\} g^{\prime}[h(x)] h^{\prime}(x) dx \), where \( h(a) = h(b) \).

Step 2: Simplify the inner function of the integral

Our focus is on the expression \( [f\{g[h(x)]\}]^{-1} f^{\prime}\{g[h(x)]\} g^{\prime}[h(x)] h^{\prime}(x) \). This can be rewritten using the derivative chain rule to resemble \( \frac{d}{dx} f(g(h(x))) \).

Step 3: Recognize the structure of the derivative

Notice that the integral structure \( [f(g(h(x)))]^{-1} f^{\prime}(g(h(x))) g^{\prime}(h(x)) h^{\prime}(x) \) matches the derivative of \( \log [f(g(h(x)))] \).

Step 4: Apply the Fundamental Theorem of Calculus

Given \( h(a) = h(b) \), the limits of the integral become equal, meaning the expression under the integral results in evaluating \( \log[f(g(h(x)))] \) from \( x=a \) to \( x=b \).

Step 5: Evaluate the integral

Since \( h(a) = h(b) \), evaluate \( \log[f(g(h(b)))] - \log[f(g(h(a)))] \), which simplifies to 0 because the upper and lower limits are the same value and hence the integral is over a degenerate interval.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a pivotal concept in integral calculus, connecting differentiation and integration, two of the main concepts in calculus. It comes in two parts, but the most relevant to our problem is the first part, which relates the evaluation of a definite integral to the antiderivative of its integrand. This theorem states that if a function is continuous on the interval \([a, b]\) and \(F\) is its antiderivative, then: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]In the step-by-step solution, the Fundamental Theorem of Calculus helps us evaluate the integral by substituting the transformation of the derivative into this framework. But importantly, because \(h(a) = h(b)\), the integral essentially spans a zero length, which brings us to consider the concept of a degenerate interval. The idea that the result is zero revolves around this special case where, effectively, the start and end points of the integral coincide.

Derivative Chain Rule

The Derivative Chain Rule is a fundamental tool for differentiating composite functions. It states that if you have a composite function \(y = f(g(x))\), then the derivative of \(y\) with respect to \(x\) is:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]In our problem, we apply the chain rule multiple times, since we are dealing with a function composed of three layers: \(f(g(h(x)))\). Thus, the derivativetransformation leads to:

• First, differentiate the external function \(f\), leading to \(f'(g(h(x)))\).
• Next, differentiate \(g\), leading to \(g'(h(x))\).
• Finally, multiply by the derivative of the most embedded function \(h\), which is \(h'(x)\).

Altogether, these form the derivative chain that we need to navigate through the composite functions. By realizing that the expression within the integral is a derivative form, connected to \(\log[f(g(h(x)))]\), it simplifies the process of integrating.

Degenerate Interval

A degenerate interval occurs when the endpoints of an interval are equal, meaning the interval has no length. In the context of integration, evaluating an integral over such an interval simplifies to zero, since there's effectively no span to sum over. In the context of our exercise, the condition \(h(a) = h(b)\) indicates that the variable does not change on the given path from \(a\) to \(b\). Thus, this turns our definite integral into one with equal bounds \(\int_{h(a)}^{h(a)} \), which conventionally resolves to 0.This concept underlines why, despite the detailed derivations and substitutions, the final value of the integral ends up being zero. It's crucial to note this, as it stresses the importance of understanding the role of the interval's limits in the determination of the integral's value.