Q 29. Calculate The Area Under The Horizontal Line Y = 4 From X = 1 To X = 3. Q 30. Find The Derivative Of The Function: Y = (x² + 2)/x.

In the realm of calculus, determining the area under a curve is a fundamental concept with vast applications in physics, engineering, and economics. When we talk about the area under a curve, we're essentially referring to the definite integral of a function over a given interval. This concept becomes particularly intuitive when dealing with simple functions like a horizontal line. In this article, we will explore how to calculate the area under a horizontal line, providing a step-by-step explanation and highlighting the underlying principles. Let's consider the specific case of finding the area under the horizontal line y = 4 from x = 1 to x = 3. This example serves as an excellent illustration of how geometric intuition and calculus can work together to solve problems.

To begin, it's crucial to visualize the problem. The equation y = 4 represents a horizontal line that intersects the y-axis at the point 4. The interval from x = 1 to x = 3 defines the boundaries within which we want to calculate the area. If we were to sketch this scenario on a coordinate plane, we would see a rectangle formed by the horizontal line y = 4, the vertical lines x = 1 and x = 3, and the x-axis. The area we are interested in is precisely the area of this rectangle. Now, recalling the basic geometric formula for the area of a rectangle – Area = Base × Height – we can easily compute the area. The base of our rectangle is the distance between the x-values 1 and 3, which is 3 - 1 = 2 units. The height of the rectangle is given by the y-value of the horizontal line, which is 4 units. Therefore, the area under the line is 2 × 4 = 8 square units. This straightforward calculation highlights the simplicity of finding the area under a horizontal line, where basic geometry provides an immediate solution. However, it's also important to understand how this relates to the concept of integration in calculus. The definite integral of a function f(x) from a to b, denoted as ∫ab f(x) dx, represents the area under the curve of f(x) between the vertical lines x = a and x = b. In our case, f(x) = 4, a = 1, and b = 3. The integral ∫13 4 dx mathematically formalizes the area calculation we performed geometrically. The antiderivative of 4 with respect to x is 4x. Evaluating this antiderivative at the limits of integration, we get 4(3) - 4(1) = 12 - 4 = 8. This result confirms our geometric calculation, showing that the area under the horizontal line y = 4 from x = 1 to x = 3 is indeed 8 square units. The harmony between geometric intuition and calculus is a powerful tool in problem-solving, and this example perfectly illustrates this synergy. By visualizing the problem and applying basic geometric principles, we can quickly arrive at the solution. Simultaneously, understanding the calculus perspective provides a deeper insight into the concept of area under a curve and its broader applications. This foundational understanding is crucial for tackling more complex problems in calculus and related fields. In summary, finding the area under a horizontal line is a straightforward process that can be approached both geometrically and through calculus. The geometric approach offers an intuitive solution based on the area of a rectangle, while the calculus approach provides a more formal method using definite integrals. Both methods, when applied correctly, lead to the same result, reinforcing the interconnectedness of different mathematical concepts.

Differentiation is a cornerstone of calculus, allowing us to determine the rate of change of a function. The derivative of a function provides valuable information about its behavior, such as its slope at any given point, its increasing or decreasing intervals, and its local maxima and minima. Mastering differentiation techniques is essential for students and professionals in various fields, including mathematics, physics, engineering, and economics. In this section, we will delve into the process of differentiating the function y = (x² + 2)/x, providing a step-by-step guide and explaining the underlying rules and principles. This particular function offers a good opportunity to apply several key differentiation techniques, including the quotient rule and the power rule. By carefully working through this example, we can solidify our understanding of differentiation and its applications.

Before we dive into the differentiation process, it's often beneficial to simplify the function if possible. In this case, we can rewrite y = (x² + 2)/x as y = x²/x + 2/x, which simplifies further to y = x + 2x⁻¹. This simplification makes the function easier to differentiate because we can now apply the power rule directly to each term. The power rule states that if f(x) = xⁿ, then its derivative f'(x) = nxⁿ⁻¹. Applying this rule to the first term, x, we have n = 1, so the derivative is 1x¹⁻¹ = 1x⁰ = 1. For the second term, 2x⁻¹, we have n = -1, so the derivative is 2(-1)x⁻¹⁻¹ = -2x⁻². Combining these results, we find that the derivative of y with respect to x is dy/dx = 1 - 2x⁻². This form of the derivative is perfectly valid, but it's often helpful to rewrite it using positive exponents and a common denominator. We can rewrite 2x⁻² as 2/x², so our derivative becomes dy/dx = 1 - 2/x². To combine these terms, we find a common denominator, which is x². Thus, we rewrite 1 as x²/x², and our derivative becomes dy/dx = x²/x² - 2/x², which simplifies to dy/dx = (x² - 2)/x². This is the simplified form of the derivative. Alternatively, if we chose not to simplify the original function, we could have applied the quotient rule directly. The quotient rule states that if y = u(x)/v(x), then dy/dx = (v(x)u'(x) - u(x)v'(x))/(v(x))². In our case, u(x) = x² + 2 and v(x) = x. The derivatives of u(x) and v(x) are u'(x) = 2x and v'(x) = 1, respectively. Applying the quotient rule, we get dy/dx = (x(2x) - (x² + 2)(1))/x². Simplifying the numerator, we have 2x² - x² - 2 = x² - 2. Thus, the derivative is dy/dx = (x² - 2)/x², which is the same result we obtained by simplifying the function first. This consistency underscores the robustness of differentiation rules and the flexibility in applying them. The derivative dy/dx = (x² - 2)/x² tells us how the function y = (x² + 2)/x changes as x changes. We can use this derivative to find critical points, intervals of increasing and decreasing behavior, and concavity. For instance, setting the derivative equal to zero, (x² - 2)/x² = 0, gives us x² - 2 = 0, so x = ±√2. These are critical points where the function may have local maxima or minima. Analyzing the sign of the derivative in different intervals can reveal whether the function is increasing or decreasing. Moreover, finding the second derivative would give us information about the concavity of the function. In conclusion, differentiating the function y = (x² + 2)/x involves applying either the power rule after simplification or the quotient rule directly. Both methods lead to the same derivative, dy/dx = (x² - 2)/x², which provides valuable insights into the behavior of the function. Mastering these differentiation techniques is crucial for understanding calculus and its applications in various fields.

Q 29. Area Under the Horizontal Line y = 4

Question: Find the area under the horizontal line y = 4 from x = 1 to x = 3.

Detailed Explanation: To find the area under the horizontal line y = 4 from x = 1 to x = 3, we need to understand the geometric interpretation of this problem. The region in question is a rectangle. The horizontal line y = 4 forms the top side of the rectangle, while the x-axis forms the bottom side. The vertical lines x = 1 and x = 3 form the left and right sides, respectively. The area of a rectangle is given by the formula: Area = Base × Height. In this case, the base of the rectangle is the distance between the x-values, which is 3 - 1 = 2 units. The height of the rectangle is the y-value, which is 4 units. Therefore, the area is 2 × 4 = 8 square units. We can also solve this problem using calculus. The area under the curve y = f(x) from x = a to x = b is given by the definite integral: ∫ab f(x) dx. In our case, f(x) = 4, a = 1, and b = 3. So, we need to evaluate the integral ∫13 4 dx. The antiderivative of 4 is 4x. Evaluating this antiderivative at the limits of integration, we get 4(3) - 4(1) = 12 - 4 = 8. This confirms our geometric calculation, showing that the area under the line y = 4 from x = 1 to x = 3 is indeed 8 square units. Therefore, the correct answer is C. 8 square units. This problem illustrates a fundamental concept in calculus: finding the area under a curve. For a simple function like a horizontal line, we can use basic geometry to solve the problem. However, for more complex functions, we need to use the techniques of integral calculus. Understanding both geometric and calculus approaches provides a comprehensive understanding of the concept. Visualizing the problem as a geometric shape helps in grasping the underlying principles. The area under a curve represents the cumulative effect of the function over an interval. In this case, it represents the cumulative value of y = 4 from x = 1 to x = 3. This concept has applications in various fields, including physics, engineering, and economics. For example, in physics, the area under a velocity-time graph represents the displacement of an object. In economics, the area under a marginal cost curve represents the total cost of production. Thus, understanding how to calculate the area under a curve is essential for solving real-world problems. In summary, the area under the horizontal line y = 4 from x = 1 to x = 3 is 8 square units. This can be calculated using both geometric and calculus methods. The geometric method involves recognizing the area as a rectangle and using the formula Area = Base × Height. The calculus method involves evaluating the definite integral ∫13 4 dx. Both methods lead to the same result, reinforcing the interconnectedness of different mathematical concepts. This problem serves as a foundational example for understanding more complex area calculations in calculus.

Q 30. Differentiation of the Function y = (x² + 2)/x

Question: Differentiate the function: y = (x² + 2)/x.

Detailed Explanation: To differentiate the function y = (x² + 2)/x, we can use several methods. One approach is to simplify the function first and then apply the power rule. Another approach is to use the quotient rule directly. Both methods will yield the same result, but one might be more convenient depending on the specific problem. Let's start by simplifying the function. We can rewrite y = (x² + 2)/x as y = x²/x + 2/x. This simplifies to y = x + 2x⁻¹. Now, we can differentiate each term separately using the power rule. The power rule states that if f(x) = xⁿ, then its derivative f'(x) = nxⁿ⁻¹. The derivative of x is 1, since the power of x is 1, and 1 × x¹⁻¹ = 1 × x⁰ = 1. For the term 2x⁻¹, the power is -1, so the derivative is 2 × (-1) × x⁻¹⁻¹ = -2x⁻². Thus, the derivative of y with respect to x is dy/dx = 1 - 2x⁻². To write this in a more conventional form, we can rewrite x⁻² as 1/x². So, dy/dx = 1 - 2/x². Now, let's find a common denominator to combine these terms. The common denominator is x². We rewrite 1 as x²/x². Therefore, dy/dx = x²/x² - 2/x² = (x² - 2)/x². This is one way to express the derivative. Alternatively, we can apply the quotient rule directly to the original function y = (x² + 2)/x. The quotient rule states that if y = u(x)/v(x), then dy/dx = (v(x)u'(x) - u(x)v'(x))/(v(x))². In our case, u(x) = x² + 2 and v(x) = x. The derivatives of u(x) and v(x) are u'(x) = 2x and v'(x) = 1, respectively. Applying the quotient rule, we get dy/dx = (x(2x) - (x² + 2)(1))/x². Simplifying the numerator, we have 2x² - (x² + 2) = 2x² - x² - 2 = x² - 2. Thus, dy/dx = (x² - 2)/x², which is the same result we obtained by simplifying the function first. This demonstrates that both methods are valid and lead to the same derivative. Now, let's compare our result with the given options. We have dy/dx = 1 - 2/x². The given options are:

A. 1-(2/x³) B. 2-(2/x³) C. 2+ (2/x²) D. 1-(2/x²)

Our result, 1 - 2/x², matches option D. 1-(2/x²). Therefore, the correct answer is D. 1-(2/x²). In summary, differentiating the function y = (x² + 2)/x can be done by simplifying the function first or by applying the quotient rule directly. Both methods lead to the same derivative, dy/dx = 1 - 2/x². This problem illustrates the importance of understanding different differentiation techniques and choosing the most efficient method for a given problem. Simplifying the function before differentiating can often make the process easier, but the quotient rule is a powerful tool for differentiating functions in the form of a fraction. Understanding the power rule and the quotient rule is essential for mastering differentiation in calculus. The derivative represents the instantaneous rate of change of the function. In this case, dy/dx tells us how y changes as x changes. This information can be used to find critical points, intervals of increasing and decreasing behavior, and concavity of the function.