Problem 30 - A Complete Resource Book in Mathematics for JEE Main

Question

The area above \(x\)-axis, bounded by the line \(x=4\) and the curve \(y=f(x)\), where \(f(x)=x^{2}, 0 \leq x \leq 1\) and \(f(x)=\sqrt{x}, x \geq 1\), is (A) 1 (B) 2 (C) 4 (D) 5

Step-by-Step Solution

Answer

The total area is 5, which is option (D).

Step 1: Understand the Problem

We need to find the area above the x-axis that is bounded by the curve \(y = f(x)\) and the vertical line \(x = 4\). The function \(f(x)\) is defined piecewise: \(f(x) = x^2\) for \(0 \leq x \leq 1\) and \(f(x) = \sqrt{x}\) for \(x \geq 1\). We need to integrate these two parts over their respective intervals.

Step 2: Set Up the Integrals

We will split the area into two parts corresponding to the two segments of the function. The first part is from \(x = 0\) to \(x = 1\) and uses \(f(x) = x^2\). The second part is from \(x = 1\) to \(x = 4\) and uses \(f(x) = \sqrt{x}\). Therefore, we have two integrals:1. \( \int_{0}^{1} x^2 \, dx \)2. \( \int_{1}^{4} \sqrt{x} \, dx \)

Step 3: Solve the First Integral

Compute the integral \( \int_{0}^{1} x^2 \, dx \):\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

Step 4: Solve the Second Integral

Compute the integral \( \int_{1}^{4} \sqrt{x} \, dx \):Since \( \sqrt{x} = x^{1/2} \), we have:\[ \int_{1}^{4} x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \left[ \frac{2x^{3/2}}{3} \right]_{1}^{4} \]Calculate the result:\[ = \frac{2(4^{3/2})}{3} - \frac{2(1^{3/2})}{3} = \frac{2 \times 8}{3} - \frac{2}{3} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3} \]

Step 5: Combine the Results

Add the results of the two integrals to find the total area:\[ \frac{1}{3} + \frac{14}{3} = \frac{15}{3} = 5 \]

Step 6: Select the Correct Answer

Based on the calculation, the total area is 5, which corresponds to option (D).

Key Concepts

Piecewise Functions

Piecewise functions are special kinds of mathematical functions that have different rules for different parts of their domain. They can be thought of as a set of "pieces," where each piece corresponds to a specific interval of values. In this exercise, the function is defined as a piecewise function, where:

• For the domain interval \(0 \leq x \leq 1\), the function is \(f(x) = x^2\).
• For \(x \geq 1\), the function switches to \(f(x) = \sqrt{x}\).

Understanding how piecewise functions work is crucial in calculus, as it helps us handle functions that behave differently across various sections of their domain. When solving problems that involve piecewise functions, it's essential to identify each piece and the intervals they correspond to, before applying methods like integration.

Area Under a Curve

Finding the area under a curve is a common problem in integral calculus, often representing the total accumulation of a quantity. In this example, the area refers to the portion of the plane bounded by the curve of the function and specific boundary lines, such as the x-axis and vertical lines. When dealing with functions, especially piecewise functions:

• First, identify the intervals over which each part of the function is defined.
• Next, set up integrals for each segment of the piecewise function separately.

In our problem, we have two parts of the function that must be assessed individually:

• From \(x = 0\) to \(x = 1\), we find the area under the curve \(y = x^2\).
• From \(x = 1\) to \(x = 4\), we consider the area under \(y = \sqrt{x}\).

By calculating these areas and summing them up, we obtain the total area under the curve between the specified boundaries.

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration, which is essentially finding the accumulation of quantities. Integration is utilized in finding areas, volumes, central points, and many other concepts in mathematics and physics. In the context of this problem, integration helps us to determine the area beneath a curve relative to the x-axis. Here's how the integration is done:

• Each piece of the function is integrated over its defined interval.
• The process involves computing the antiderivative, setting up definite integrals, and evaluating them over their specified limits.

For the piece \(f(x) = x^2\):

• The area is \(\int_{0}^{1} x^2 \, dx\), resulting in \(\frac{1}{3}\).

For the piece \(f(x) = \sqrt{x}\):

• The integral is \(\int_{1}^{4} \sqrt{x} \, dx\) which evaluates to \(\frac{14}{3}\).

Finally, to find the total area, we simply add the two resulting areas. This shows the importance of properly applying integral calculus to explore and compute areas involving complex functions.