Problem 190 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)\) be a function satisfying \(f^{\prime}(x)=f(x)\) with \(f(0)=1\) and \(g(x)\) be a function that satisfies \(f(x)+g(x)\) \(=x^{2}\). Then the value of the integral \(\int_{0}^{1} f(x) g(x) d x\) is (A) \(e-\frac{e^{2}}{2}-\frac{5}{2}\) (B) \(e+\frac{e^{2}}{2}-\frac{3}{2}\) (C) \(e-\frac{e^{2}}{2}-\frac{3}{2}\) (D) \(e+\frac{e^{2}}{2}+\frac{5}{2}\)

Step-by-Step Solution

Answer

The value of the integral is option (C): \( e - \frac{1}{2} e^2 - \frac{3}{2} \).

Step 1: Identify f(x) from Differential Equation

Given that \( f'(x) = f(x) \) and \( f(0) = 1 \), this is a classic differential equation whose solution is \( f(x) = Ce^x \). Using the initial condition, \( f(0) = C \cdot e^0 = 1 \), we find \( C = 1 \), so \( f(x) = e^x \).

Step 2: Express g(x) Using Given Relation

We are also given that \( f(x) + g(x) = x^2 \). Since \( f(x) = e^x \), we can express \( g(x) \) as \( g(x) = x^2 - e^x \).

Step 3: Set Up the Integral

We need to evaluate the integral \( \int_{0}^{1} f(x)g(x) \, dx \). Substituting \( f(x) = e^x \) and \( g(x) = x^2 - e^x \), this becomes \( \int_{0}^{1} e^x (x^2 - e^x) \, dx \).

Step 4: Distribute and Split the Integral

Distribute within the integral: \( \int_{0}^{1} e^x (x^2 - e^x) \, dx = \int_{0}^{1} e^x x^2 \, dx - \int_{0}^{1} e^{2x} \, dx \).

Step 5: Evaluate e^x x^2 Term

To find \( \int_{0}^{1} e^x x^2 \, dx \), use integration by parts. Let \( u = x^2 \) and \( dv = e^x dx \), so \( du = 2x dx \) and \( v = e^x \). Applying integration by parts formula, we have \( \int u \, dv = uv - \int v \, du \). This results in a multi-step calculation.

Step 6: Integration by Parts for e^x x^2

Apply integration by parts: \( \left. x^2 e^x \right|_0^1 - \int_0^1 2x e^x \, dx = \left( e \cdot 1^2 - (0) \right) - 2 \int_0^1 x e^x \, dx \). Now solve \( \int x e^x \, dx \) using integration by parts again: \( u = x, dv = e^x dx \), so \( du = dx, v = e^x \).

Step 7: Second Integration by Parts

Repeat: \( \int x e^x \, dx = \left. x e^x \right|_0^1 - \int_{0}^{1} e^x \, dx = \left( e-0 \right) - (e - 1) = 1.\) So, \( \int_{0}^{1} e^x x^2 \, dx = e - 2(1) = e - 2. \)

Step 8: Evaluate e^{2x} Term

Find \( \int_{0}^{1} e^{2x} \, dx \). The antiderivative is \( \frac{1}{2} e^{2x} \), and evaluating from 0 to 1 gives \( \frac{1}{2} (e^2 - 1) = \frac{1}{2} e^2 - \frac{1}{2}.\)

Step 9: Combine Results

Combine both parts: \( (e - 2) - (\frac{1}{2} e^2 - \frac{1}{2}) = e - 2 - \frac{1}{2} e^2 + \frac{1}{2} \). Simplifying gives \( e - \frac{1}{2} e^2 - \frac{3}{2}.\)

Step 10: Choose the Correct Option

According to our calculation, the correct expression for the integral is \( e - \frac{1}{2} e^2 - \frac{3}{2} \), corresponding to option (C).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In simple terms, they tell us how a function changes over time or space. They are fundamental in modeling real-world phenomena, from physics and engineering to biology and economics.

In this exercise, the differential equation given is \( f'(x) = f(x) \). This is a first-order differential equation with a classic form, where the rate of change of \( f(x) \) is directly proportional to the function itself. The solution to such equations can usually be written in the form of an exponential function.

• The exponential solution means \( f(x) = Ce^x \), where \( C \) is a constant determined by initial conditions.
• To find \( C \), substitute the initial condition \( f(0) = 1 \) into the equation. This gives \( C = 1 \), so the specific solution here is \( f(x) = e^x \).

Differential equations often need initial conditions, like \( f(0) = 1 \), to find a specific solution that is not just a general formula.

Integration by Parts

Integration by parts is a powerful technique used in integral calculus to solve integrals involving products of functions. It's based on the product rule for differentiation and helps convert complex integrals into simpler, more manageable ones.

The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]This exercise involves using integration by parts to evaluate the integral of \( e^x x^2 \). Here's how you typically set it up:

• Choose \( u \) and \( dv \) wisely. For \( e^x x^2 \), let \( u = x^2 \) and \( dv = e^x dx \).
• This implies \( du = 2x dx \) and \( v = e^x \).
• Apply the formula: \( \int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx \).

In complicated integrals, you might need to repeat the integration by parts process, like when you find \( \int x e^x \, dx \) using integration by parts again.

Initial Conditions

Initial conditions are crucial when solving differential equations because they help specify a particular solution from a family of solutions. In mathematical context, an initial condition gives the value of the function (and possibly its derivatives) at a specific point.

In our exercise, the initial condition \( f(0) = 1 \) is used. This condition helps us find the constant \( C \) in the general solution of the differential equation \( f(x) = Ce^x \).

• Without an initial condition, the solution \( f(x) = Ce^x \) remains undetermined, as \( C \) could be any constant.
• By substituting \( f(0) = 1 \), we can solve for \( C \) and make the solution specific and relevant to the problem at hand.

Initial conditions essentially "anchor" the solution, ensuring it accurately reflects the scenario or phenomenon being modeled.