Problem 108 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{0}^{x+y} e^{\sin ^{2} t} d t-\int_{0}^{y} e^{\sin ^{2} t} d t\right]\), where \(y\) is a constant independent of \(x\), is equal to (A) \(e^{\sin ^{2} y}\) (B) \(2 e^{\sin ^{2} y}\) (C) \(-e^{\sin ^{2} y}\) (D) None of these

Step-by-Step Solution

Answer

The limit is equal to \(e^{\sin ^{2} y}\), so the answer is (A).

Step 1: Rewriting the Expression

Start by rewriting the expression from the problem. We have:\[\lim_{x \to 0} \frac{1}{x} \left[ \int_{0}^{x+y} e^{\sin^2 t} \, dt - \int_{0}^{y} e^{\sin^2 t} \, dt \right]\]This can be rewritten by combining the integrals as:\[\frac{1}{x} \int_{y}^{x+y} e^{\sin^2 t} \, dt\]This simplification is possible because the fundamental theorem of calculus allows us to express the difference between two integrals from 0 to different bounds as an integral with limits being those bounds.

Step 2: Using The Substitute Technique

Introduce a substitution to simplify the limits of the integral. Let \( u = t-y \), which implies that \( du = dt \). When \( t = y \), \( u = 0 \) and when \( t = x+y \), \( u = x \). Thus, the integral becomes:\[\int_{0}^{x} e^{\sin^2(u+y)} \, du\]The rewritten expression is:\[\lim_{x \to 0} \frac{1}{x} \int_{0}^{x} e^{\sin^2(u+y)} \, du\]

Step 3: Applying the Mean Value Theorem for Integrals

Applying the Mean Value Theorem for integrals, which states that for a continuous function \( f \) on \([a, b]\), there exists a \( c \in [a, b] \) such that:\[\int_{a}^{b} f(t) \, dt = (b-a)f(c)\]Set \( f(u) = e^{\sin^2(u+y)} \), hence:\[\int_{0}^{x} e^{\sin^2(u+y)} \, du = x \cdot e^{\sin^2(c+y)}\]where \( c \in [0, x] \). The expression becomes:\[\lim_{x \to 0} e^{\sin^2(c+y)} = e^{\sin^2(y)}\]since as \( x \to 0 \), \( c \to 0 \).

Step 4: Evaluating the Limit

Thus, evaluating the limit yields:\[\lim_{x \to 0} \frac{1}{x} \cdot x \cdot e^{\sin^2(c+y)} = e^{\sin^2(y)}\]We have applied the Mean Value Theorem for integrals correctly, leading to elucidating that the limit evaluates precisely to the function value at \( y \).

Key Concepts

integrals and derivatives

In calculus, integrals and derivatives are fundamental concepts that describe the accumulation and rate of change in mathematical functions.
Derivatives represent the rate at which a function changes at any given point. They are closely linked to the concept of a slope in geometry, except they can measure the instantaneous rate of change.
On the other hand, integrals are used to calculate the area under a curve, which represents accumulation or total size. Integrals can be considered as the reverse operation of derivatives.

• Integrals accumulate quantities over intervals.
• Derivatives measure rates of change.

In the exercise, the problem revolves around the manipulation and evaluation of integrals using the limit as it approaches zero. This closely ties the relationship between integrals and derivatives, as the limit reflects how integrals behave similarly to derivatives in certain contexts.

fundamental theorem of calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration. It consists of two main parts:
The first part states that if you have a continuous function and you take its integral over an interval, then the derivative of this integral is the original function.

• It explains why differentiation and integration are inverse processes.
• It allows for the computation of definite integrals through antiderivatives.

In the context of the exercise, the fundamental theorem allows us to manipulate the integral expressions to combine them, streamlining the problem into a form that is easier to solve with limits.

mean value theorem

The Mean Value Theorem (MVT) in calculus is an important tool that provides a connection between the average rate of change of a function and its instantaneous rate of change. Specifically for integrals, the Mean Value Theorem states that if a function is continuous over an interval, there exists at least one point within that interval where the function's value is equal to the average value of the function over the interval.
This is expressed as:

• \[\int_{a}^{b} f(t) \, dt = (b-a)f(c)\]
• where \( c \) is within the interval \([a, b]\).

In the exercise, the Mean Value Theorem was applied to transform the integral, helping us express it in terms of a specific function value, making it easier to evaluate the limit. This highlights the understanding that even complex integral expressions can have simpler meanings when evaluated under specific conditions provided by the theorem.