Problem 108 - A Complete Resource Book in Mathematics for JEE Main 2017

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 solution is (A) \(e^{\sin^2 y}\).

Step 1: Understand the Expression

We are given the limit expression \( \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{0}^{x+y} e^{\sin ^{2} t} d t-\int_{0}^{y} e^{\sin ^{2} t} d t\right] \). Here, the two integrals are defined over different bounds. The expression inside the limit is a form of difference quotient, resembling a derivative.

Step 2: Simplify the Expression

Notice that the expression inside the limit \( \frac{1}{x} \left[ F(x+y) - F(y) \right] \) is of the form of a derivative, where \( F(t) = \int_{0}^{t} e^{\sin^2 u} du \). First, apply the Fundamental Theorem of Calculus to express \( F(t) \) as: \[ F(t) = \int_{0}^{t} e^{\sin^2 u} du \Rightarrow F'(t) = e^{\sin^2 t}. \]

Step 3: Apply the Fundamental Theorem of Calculus

Given the form of derivative, apply the derivative to \( F(x+y) \):\[ \lim_{x \to 0} \frac{1}{x} \left[ F(x+y) - F(y) \right] = F'(y). \]Since \( F'(t) = e^{\sin^2 t} \), replace \( t \) by \( y \):\[ F'(y) = e^{\sin^2 y}. \]

Step 4: Solve the Limit

Now substitute \( F'(y) = e^{\sin^2 y} \) into the limit expression, based on our calculus approach:\[ \lim_{x \to 0} \frac{1}{x}\left[ F(x+y) - F(y) \right] = e^{\sin^2 y}. \]

Step 5: Confirm Final Result

With the solution oriented around finding the derivative, the calculated limit aligns with the option (A), which is \( e^{\sin^2 y} \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is like a bridge that connects the world of derivatives with the world of integrals. It's a powerful tool in calculus that helps us evaluate definite integrals using antiderivatives.

• Part 1: If you have a continuous function, its integral gives you another function, which we call the antiderivative. In simple terms, an antiderivative is a function whose derivative is the original function.
• Part 2: This part tells us how to evaluate definite integrals. If you have a function which is an antiderivative of your original function, then to find the integral from point \(a\) to \(b\), you just find the difference \(F(b) - F(a)\).

In this exercise, we use the Fundamental Theorem to express the function \(F(t) = \int_{0}^{t} e^{\sin^2 u} \, du\) and its derivative \(F'(t) = e^{\sin^2 t}\). This makes the limit expression clearer and facilitates solving the problem.

Difference quotient

The difference quotient is a crucial concept when discussing limits and derivatives. It serves as a stepping stone to finding the derivative of a function, allowing us to measure how a function changes as its input changes by a tiny amount.
In mathematical terms, the difference quotient is expressed as \( \frac{f(x+h) - f(x)}{h} \), where \( h \) is the small change in \( x \). As \( h \) tends to zero, this quotient gives us the derivative at that point.

• Form: The exercise shows a limit of the form \( \frac{1}{x} [F(x+y) - F(y) ]\). This resembles the difference quotient notation, indicating we are evaluating the rate of change of \( F(t) \) at \( t = y \).
• Application: Here, as \( x \rightarrow 0 \), it helps us find the derivative \( F'(y) \), which simplifies our expression to \( e^{\sin^2 y} \).

Understanding this concept is vital for activities like taking derivatives and working with limits, and it connects directly to our task of simplifying the expression in the challenge.

Derivative of an integral

The derivative of an integral is a concept involving taking the derivative of an expression that's calculated using an integral. It might sound complex, but it's very simple when broken down using the Fundamental Theorem of Calculus.
Simply put, taking the derivative of an integral basically brings us back to the original function.

• The theorem part we exploit here is that if \(F(t) = \int_{0}^{t} e^{\sin^2 u} \, du\), then \(F'(t) = e^{\sin^2 t} \).
• In the problem, this allows us to transform our integral expression within the limit into something we can differentiate easily.
• By recognizing the form \( F(x+y) - F(y) \), the derivative \( F'(t) \) simplifies the process to directly evaluate \( F'(y) = e^{\sin^2 y} \).

Using these steps seamlessly integrates the derivative and integral process, essential for evaluating and simplifying limit expressions like the one in the exercise.