Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(y)=e^{y}, g(y)=y ; y>0\) and \(\phi(t)=\int_{0}^{t} f(t-y) g(y) d y\), then \(\phi(t)=\) (A) \(e^{t}-(1+t)\) (B) \(1-e^{-t}(1+t)\) (C) \(t e^{t}\) (D) None of these

Step-by-Step Solution

Answer

(B) \( 1 - e^{-t}(1 + t) \)

Step 1: Substitute Functions into Integral

The given integral is \( \phi(t) = \int_{0}^{t} f(t-y) g(y) \, dy \). First, substitute the functions \( f(y) = e^y \) and \( g(y) = y \) into the integral. This gives us the expression \( \phi(t) = \int_{0}^{t} e^{t-y} \cdot y \, dy \).

Step 2: Simplify the Integral

Simplify the expression inside the integral: \( e^{t-y} \cdot y = e^t \cdot y \cdot e^{-y} \). This rewrites the integral as \( \phi(t) = e^t \int_{0}^{t} y e^{-y} \, dy \).

Step 3: Integrate by Parts

To solve the integral \( \int_{0}^{t} y e^{-y} \, dy \), use integration by parts where \( u = y \) and \( dv = e^{-y} \, dy \). Thus, \( du = dy \) and \( v = -e^{-y} \). The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \].

Step 4: Calculate the Parts

Compute \( uv = -y e^{-y} \) evaluated from 0 to \( t \), which gives \( -t e^{-t} + 0 \). Then compute \( \int v \cdot du = \int -e^{-y} \, dy \), which simplifies to \( e^{-y} \) evaluated from 0 to \( t \): \( e^{-0} - e^{-t} = 1 - e^{-t} \).

Step 5: Combine the Results

From integration by parts, we found:\[ \int_{0}^{t} y e^{-y} \, dy = -t e^{-t} + (1 - e^{-t}) = 1 - e^{-t}(1 + t) \] .Therefore, \( \phi(t) = e^{t}(1 - e^{-t}(1 + t)) = 1 - e^{-t}(1 + t) \).

Step 6: Determine the Correct Answer

Comparing with the provided options, \( \phi(t) = 1 - e^{-t}(1 + t) \) matches option (B). Thus, the correct answer is (B) \( 1 - e^{-t}(1 + t) \).

Key Concepts

Integration by Parts

Integration by Parts is a technique used to solve integrals involving a product of two functions. Recall that the formula for this method is \[ \int u \, dv = uv - \int v \, du \]. When choosing parts, one function is assigned to \( u \) and the other to \( dv \). A common strategy is to let \( u \) be a function that becomes simpler when differentiated and \( dv \) a function that remains simple when integrated. This approach helps to break down the integral into manageable pieces.

• Let \( u = y \), then \( du = dy \).
• Let \( dv = e^{-y} \, dy \), then \( v = -e^{-y} \).

Once functions are chosen, apply the formula by evaluating \( uv \) and the remaining integral \( \int v \, du \) over the specified limits. The key is to end with an integral that is easier to solve than the original.

Exponential Functions

Exponential functions involve the constant \( e \) (approximately equal to 2.71828), raised to a variable power. They are unique due to their continuous growth or decay. An essential property is that the derivative of \( e^x \) is itself, maintaining the function's form under differentiation and integration.
In this problem, \( f(y) = e^y \), but since it's written as \( f(t-y) \) in the integral, it is transformed into \( e^{t-y} = e^t \cdot e^{-y} \). This separation allows the \( e^t \) factor to be brought outside the integral, simplifying the expression and focusing on the integral of \( y \, e^{-y} \).
When dealing with exponentials within integrals, remembering their differentiability and integrability helps manage complex expressions more easily.

Definite Integrals

Definite integrals calculate the total accumulation of a quantity, defined over a specified interval, here from \( 0 \) to \( t \). This range specifies the bounds and limits of integration. Resulting values from this process often represent areas under a curve.
In the given exercise, solving the definite integral \( \int_{0}^{t} y \, e^{-y} \, dy \) involves finding the antiderivative and then evaluating it at these bounds. The integration by parts technique simplifies this definite integral.
Finally, after performing the steps of integration by parts, including evaluating the antiderivatives at the limits, the answer is obtained by substituting back the computed results. This process illustrates how definite integrals are used not just to find solutions, but to verify them within specified limits as described in the answer to option \( (B) \).