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

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

The answer is (A) \(e^t - (1 + t)\).

Step 1: Understand the function

First, recognize that the function \(\phi(t)\) involves an integral of the product of two functions, \(f(t-y)\) and \(g(y)\). Here, \(f(y) = e^y\) and \(g(y) = y\).

Step 2: Simplify the expression inside the integral

Since we have \(f(t-y) = e^{t-y}\) and \(g(y) = y\), the integral becomes \(\phi(t) = \int_{0}^{t} e^{t-y} y \, dy\).

Step 3: Change variables to simplify the integral

To make integration easier, let \(u = t-y\), which implies \(du = -dy\). The limits of integration change from \(y=0\) to \(y=t\) when \(u=t\) to \(u=0\), so the integral becomes \(-\int_{t}^{0} e^u (t-u) (-du) \), which simplifies to \(\int_{0}^{t} e^u (t-u) \, du\).

Step 4: Split the integral

Distribute the \(e^u\) to get two integrals: \(\int_{0}^{t} t e^u \, du - \int_{0}^{t} u e^u \, du\).

Step 5: Compute the first integral

Evaluate \(\int_{0}^{t} t e^u \, du = t \int_{0}^{t} e^u \, du = t[e^u]_{0}^{t} = te^t - t\).

Step 6: Compute the second integral by parts

Use integration by parts for \(\int_{0}^{t} u e^u \, du\) where \(u = u\) and \(dv = e^u du\). Let \(v = e^u\) and \(du = du\). Then, \(\int u e^u \, du = ue^u - \int e^u \, du\). Substituting back gives \([ue^u - e^u]_{0}^{t} = te^t - e^t + 1\).

Step 7: Combine results

Combine the two results: \(te^t - t - (te^t - e^t + 1) = te^t - t - te^t + e^t - 1 = e^t - 1 - t\). Therefore, the correct answer is \(e^t - (1 + t)\).

Step 8: Conclusion: Match with options

After simplifying the expression, \(\phi(t) = e^t - (1 + t)\), which corresponds to option (A).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to integrate the product of two functions. It is derived from the product rule of differentiation and is generally used when an integral cannot be easily evaluated using basic techniques. The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
To apply this formula, follow these steps:

• Identify two components within the integral: one part of the function (\(u\)) to differentiate and another part (\(dv\)) to integrate.
• Differentiate \(u\) to get \(du\), and integrate \(dv\) to get \(v\).
• Substitute these into the integration by parts formula.
• Simplify and evaluate the remaining integral if necessary.

This method is particularly helpful when dealing with integrals like \(\int u e^u \, du\), as seen in the original exercise solution. It simplifies complex expressions by transforming them into simpler parts, allowing the integration to proceed step-by-step.

Definite Integrals

Definite integrals are used to calculate the accumulated total of a function over a specific interval. In contrast to indefinite integrals, which produce a general function plus a constant (\(C\)), definite integrals result in a numerical value that represents the area under the curve of the function between two specified limits. The notation for a definite integral is:
\[ \int_{a}^{b} f(x) \, dx \]
This computes the total area under the curve \(f(x)\) from \(x = a\) to \(x = b\). Here are some key points about definite integrals:

• The Fundamental Theorem of Calculus connects differentiation and integration, enabling us to evaluate a definite integral by finding an antiderivative of the function.
• The result of a definite integral can be interpreted as the net area, taking into account any area below the x-axis as negative.
• In practice, definite integrals are crucial in physics and engineering for calculating physical quantities like work, energy, and probabilities.

In the exercise, the definite integral \(\int_{0}^{t} e^{u}(t-u) \, du\) is evaluated, dividing into simpler parts to find the exact value of the area under that function from 0 to \(t\).

Exponential Functions

Exponential functions, especially those involving \(e\), occur frequently in calculus because of their unique properties of growth or decay. The function \(e^x\) is special due to the fact that it is its own derivative, making it particularly significant in calculus. Important characteristics of exponential functions include:

• Rapid growth/decay characterized by the base \(e \, (\approx 2.718)\), which is a key mathematical constant.
• The derivative of \(e^x\) is itself, \(\frac{d}{dx}e^x = e^x\), which simplifies many calculus operations.
• Exponential functions can model real-world phenomena such as population growth, radioactive decay, and compound interest.

In the context of the original problem, the exponential function \(e^{t-y}\) appears within an integral. This shows how such functions are used in combination with other calculus techniques, like integration by parts, to evaluate more complex expressions. Understanding how boundary values affect exponential functions within definite integrals is an essential skill.