NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Mathematics plays a crucial role in the academic journey of a Class 12 student, especially for those preparing for board exams and competitive exams like JEE Main and Advanced. Chapter 8 of the NCERT Mathematics textbook, "Application of Integrals," is one of the significant chapters that provide students with comprehensive insights into applying integrals in various real-world problems. This chapter is vital not just for exams but also for understanding the mathematical concepts that can be applied in fields like physics, engineering, and economics.

This article will provide a detailed overview of Chapter 8, covering the main topics, important formulas, solved examples, and the significance of practicing NCERT solutions for excelling in exams.

Understanding Integrals and Their Applications

Integrals are an essential part of calculus, a branch of mathematics dealing with continuous change. While differentiation is concerned with the rate of change, integration is all about accumulation. In simpler terms, integration helps in finding areas under curves, volumes, and other quantities that arise when summing infinitesimal data points.

In Chapter 8, the primary focus is on applying integrals to calculate areas under curves and between curves. Understanding this application is crucial for solving problems that involve finding lengths, areas, and volumes in various contexts.

Key Topics Covered in Chapter 8: Application of Integrals

1. Introduction to Definite Integrals
2. Area Under Simple Curves
3. Area Between Two Curves

Let’s discuss these topics in detail:

1. Introduction to Definite Integrals

Before diving into the application of integrals, it's essential to understand definite integrals. A definite integral is represented as:

$$\int_{a}^{b} f(x) \, dx$$

It provides the area under the curve of $f(x)$ from point $x = a$ to $x = b$. In this chapter, students learn about the properties of definite integrals and their significance in calculating the area under a curve.

2. Area Under Simple Curves

One of the primary applications of integrals is to find the area under a curve. For a given function $y = f(x)$, the area between the curve and the x-axis from $x = a$ to $x = b$ can be calculated using the formula:

$$\text{Area} = \int_{a}^{b} f(x) \, dx$$

For example, to find the area under the curve of $y = x^2$ from $x = 0$ to $x = 3$:

$$\text{Area} = \int_{0}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{27}{3} - 0 = 9$$

3. Area Between Two Curves

The chapter also focuses on finding the area between two curves. For two curves $y = f(x)$ and $y = g(x)$ where $f(x) \geq g(x)$ for all $x$ in the interval $[a, b]$, the area between the two curves is given by:

$$\text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx$$

For example, consider finding the area between the curves $y = x^2$ and $y = 4x - x^2$ from $x = 0$ to $x = 2$. First, we determine the points of intersection by equating the two curves:

$$x^2 = 4x - x^2 \implies 2x^2 - 4x = 0 \implies x(x - 2) = 0$$

So, the points of intersection are $x = 0$ and $x = 2$. The required area can be calculated as:

$$\text{Area} = \int_{0}^{2} [(4x - x^2) - x^2] \, dx = \int_{0}^{2} (4x - 2x^2) \, dx$$ $$\text{Area} = \left[ 2x^2 - \frac{2x^3}{3} \right]_{0}^{2} = (8 - \frac{16}{3}) - (0) = \frac{8}{3} \, \text{square units.}$$

Importance of Practicing NCERT Solutions for Chapter 8

The NCERT Solutions for Class 12 Maths Chapter 8 provide step-by-step explanations and solutions to all the exercises. Practicing these solutions is crucial for several reasons:

1. Conceptual Clarity: The NCERT solutions are designed to help students grasp the underlying concepts of integrals and their applications. This clarity is crucial for solving complex problems that involve multiple steps and require a deep understanding of calculus.

2. Board Exam Preparation: For Class 12 students, scoring high in board exams is essential. The questions in board exams often directly come from NCERT textbooks or are based on similar patterns. Practicing NCERT solutions ensures that students are well-prepared for any type of question related to integrals that may appear in their exams.

3. Foundation for Competitive Exams: Many competitive exams like JEE Main and Advanced require a solid understanding of calculus, including integrals. Practicing NCERT solutions helps build a strong foundation, making it easier to tackle higher-level problems in these exams.

4. Problem-Solving Skills: Mathematics is not just about memorizing formulas but also about applying them correctly in different contexts. The variety of problems provided in NCERT solutions helps enhance problem-solving skills, analytical thinking, and the ability to apply concepts in different scenarios.

5. Boosting Confidence: Consistent practice of NCERT solutions helps boost a student’s confidence, reducing exam anxiety and helping them approach problems more systematically.

Important Formulas from Chapter 8: Application of Integrals

Here are some important formulas that are frequently used in Chapter 8:

1. Definite Integral as the Limit of a Sum:

   $$\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i) \Delta x$$

2. Area Under a Curve:

   $$\text{Area} = \int_{a}^{b} f(x) \, dx$$

3. Area Between Two Curves:

   $$\text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx, \text{ where } f(x) \geq g(x) \text{ in } [a, b]$$

4. Properties of Definite Integrals:

   • $\int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx$
   • $\int_{a}^{a} f(x) \, dx = 0$
   • $\int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx$

Step-by-Step Solved Examples from NCERT Solutions

Here are a few solved examples that illustrate the application of integrals:

Example 1: Find the area enclosed by the parabola $y = x^2$ and the line $y = 4x$.

Solution:

1. First, find the points of intersection by equating the two equations:

   $$x^2 = 4x \implies x(x - 4) = 0$$

   The points of intersection are $x = 0$ and $x = 4$.

2. Set up the integral to find the area:

   $$\text{Area} = \int_{0}^{4} (4x - x^2) \, dx$$

3. Solve the integral:

   $$\text{Area} = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = (32 - \frac{64}{3}) - (0) = \frac{32}{3} \, \text{square units.}$$

Example 2: Calculate the area bounded by the curve $y = \sqrt{x}$, the x-axis, and the lines $x = 1$ and $x = 4$.

Solution:

1. Set up the integral for the area:

   $$\text{Area} = \int_{1}^{4} \sqrt{x} \, dx$$

2. Solve the integral:

   $$\text{Area} = \left[ \frac{2x^{3/2}}{3} \right]_{1}^{4} = \left( \frac{2(8)}{3} \right) - \left( \frac{2(1)}{3} \right) = \frac{14}{3} \, \text{square units.}$$

Conclusion

Chapter 8 of the NCERT Class 12 Maths textbook, "Application of Integrals," is vital for students aiming to score well in their exams and build a solid foundation in calculus. The chapter provides a comprehensive understanding of applying integrals to calculate areas under curves and between curves. Mastering these concepts is crucial not only for board exams but also for competitive exams like JEE Main and Advanced.

By practicing the NCERT solutions for this chapter, students can gain a deeper understanding of the concepts, enhance their problem-solving skills, and boost their confidence in handling complex mathematical problems. Integrals are a gateway to understanding many real-world applications, making this chapter a must-study for every aspiring mathematician, engineer, or scientist.