Problem 184 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(y)=e^{y}, g(y)=y ; y>0\) and \(F(t)=\int_{0}^{t} f(t-y) g(y) d y\), then \([\mathbf{2 0 0 3}]\) (A) \(F(t)=1-e^{-t}(1+t)\) (B) \(F(t)=e^{t}-(1+t)\) (C) \(F(t)=t e^{t}\) (D) \(F(t)=t e^{-t}\)

Step-by-Step Solution

Answer

The correct answer is (A) \(F(t)=1-e^{-t}(1+t)\).

Step 1: Identify the Given Functions

We are given the functions \(f(y) = e^y\) and \(g(y) = y\). We also have \(F(t) = \int_{0}^{t} f(t-y) g(y) \, dy\). Our task is to find \(F(t)\).

Step 2: Substitute into the Integral

Substitute \(f(t-y) = e^{t-y}\) and \(g(y) = y\) into the integral: \[F(t) = \int_{0}^{t} e^{t-y} \cdot y \, dy\].

Step 3: Simplify the Expression

Rearrange the integral: \[F(t) = e^t \int_{0}^{t} y e^{-y} \, dy\]. Notice that \(e^t\) is a constant with respect to \(y\), so it can be taken outside of the integral.

Step 4: Solve the Integral

Recognize the integral: \(\int y e^{-y} \, dy\). Use integration by parts where \(u = y\) and \(dv = e^{-y} dy\). We get \(du = dy\) and \(v = -e^{-y}\). Applying integration by parts gives: \[\int y e^{-y} \, dy = -y e^{-y} - \int -e^{-y} \, dy = -y e^{-y} + e^{-y} + C\].

Step 5: Evaluate Definite Integral

Evaluate the integral from 0 to \(t\): \[\left[-y e^{-y} + e^{-y} \right]_{0}^{t} = \left[-t e^{-t} + e^{-t} \right] - \left[0 + 1\right]\].

Step 6: Apply Boundaries and Simplify

Calculate: \(-t e^{-t} + e^{-t} - 1 = e^{-t} (1 - t) - 1\). Factor out \(e^{-t}\): \[= 1 - e^{-t}(1 + t)\].

Step 7: Determine the Correct Answer

The expression calculated: \(1 - e^{-t}(1 + t)\) matches with option (A). Thus, \(F(t) = 1 - e^{-t}(1 + t)\).

Key Concepts

Exponential Functions

Exponential functions are one of the most important functions in mathematics, characterized by a constant base raised to a variable exponent. The most well-known exponential function involves the base "e," approximately 2.718. This base is known as Euler's number and is fundamental in calculus and many natural phenomena. The function

• An exponential function can be expressed as \( f(x) = e^x \), where each increase in \( x \) results in a proportional increase in \( f(x) \).
• These functions are notably continuous and smooth; they never break or have abrupt changes.
• Additionally, exponential functions grow very rapidly, which can model processes like population growth or radioactive decay.

An interesting feature of these functions is that the rate of growth of the function is directly proportional to its current value. This means that as the value grows, the rate of increase also accelerates. When solving integrals or derivatives involving exponential functions, the unique properties allow for simplifying calculations using specific methods like integration by parts, observed in the provided exercise.

Integration by Parts

Integration by parts is a powerful method for solving integrals, specifically when the product of two functions is involved. This technique is derived from the product rule of differentiation and is invaluable in integrating products such as the exponential function multiplied by a polynomial function.The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]Key to success with this technique is choosing the correct functions for \( u \) and \( dv \).

• Typically, \( u \) is chosen to be a function that becomes simpler when differentiated, in our case \( u = y \).
• \( dv \) should be a function that is easy to integrate, here \( dv = e^{-y} \ dy \).
• Differentiating \( u \), we get \( du = dy \) and integrating \( dv \), we find \( v = -e^{-y} \).

Substituting into the formula yields a simplified expression that can be further evaluated to solve the integral. This method is particularly useful when faced with the task of evaluating definite integrals, where results need to be calculated between specific limits, as demonstrated in the solution steps.

Function Composition

Function composition involves creating a new function by applying one function to the results of another. It's a fundamental concept when dealing with complex functions, enabling us to simplify and solve problems that involve multiple layers of function application.Consider the scenario from the exercise where we have:- \( f(t-y) = e^{t-y} \)- \( g(y) = y \)This creates the integral \( F(t) = \int_0^t f(t-y)g(y) \, dy \), a classic example of function composition. Here, the values from function "g" are input into function "f," constructing a compounded result.

• When composing functions, the order matters since the output of one becomes the input of another.
• This technique helps in breaking down complex expressions into manageable pieces that can be analyzed independently.
• In calculus, understanding and leveraging function composition is critical for analyzing integrals like the one given, where transformations simplify integration.

By setting up the expression correctly and identifying each function's role in the composition, we can better tackle intricate integrals and extract solutions effectively. This organized approach provides clarity when solving problems involving multiple variables and operations.