CBSE Class 12 Mathematics Chapter 8: Applications of Integrals NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for Chapter 8 of the CBSE Class 12 Mathematics syllabus, focusing on the Applications of Integrals. It covers finding the area of various regions bounded by curves, lines, and axes, particularly in the first quadrant. The solutions explain the process of setting up definite integrals based on the given boundaries and evaluating them to determine the area. Key concepts include understanding the relationship between the curve equation and the area, and correctly applying integration techniques. These solutions are designed to help students grasp the fundamental principles of calculating areas using integration, offering step-by-step guidance for each problem. They serve as an excellent tool for exam preparation, enabling students to practice and reinforce their understanding of this important calculus topic.

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Quick info

Board	CBSE
Class	Class 12
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 8

Chapter summary

Chapter 8 of the CBSE Class 12 Mathematics syllabus, Applications of Integrals, focuses on calculating areas of regions. This section provides NCERT Solutions that guide students through finding areas bounded by simple curves like parabolas, lines, and the x or y-axis. The solutions demonstrate how to set up and evaluate definite integrals to compute these areas, emphasizing the importance of correctly identifying the limits of integration and the integrand. It covers problems involving areas in the first quadrant.

Learning outcomes

Understand the concept of finding areas using definite integrals.
Identify the region bounded by given curves and lines.
Set up the correct integral expression for the area.
Evaluate definite integrals to calculate the area.
Apply integration to solve problems involving areas in the first quadrant.

Topics covered

Paper topics

Area under a curve
Area bounded by curves and lines
Definite integrals for area calculation
Integration with respect to x
Integration with respect to y
Area in the first quadrant
Parabolic curves
Linear boundaries

Important topics

Setting up integrals for area
Evaluating definite integrals
Area bounded by y^2=x and lines
Area bounded by x^2=4y and lines
Understanding integration limits

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Questions and Solutions

Question 1

Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$ , $x = 4$ , and the x-axis in the first quadrant.

Solution: The region is bounded by the curve

y^2 = x

, the vertical lines

x = 1

and

x = 4

, and the x-axis. Since we are considering the first quadrant,

y \ge 0

From $y^2 = x$ , we get $y = \sqrt{x}$ (taking the positive root for the first quadrant).

The area

A

can be calculated by integrating

y

with respect to

x

from

x = 1

x = 4

$A = \int_{1}^{4} y \, dx$

Substitute

y = \sqrt{x}

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

Now, evaluate the integral:

$A = \left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}$

$A = \frac{2}{3} \left[x^{\frac{3}{2}}\right]_{1}^{4}$

Substitute the limits of integration:

$A = \frac{2}{3} \left[\left(4\right)^{\frac{3}{2}} - \left(1\right)^{\frac{3}{2}}\right]$

$A = \frac{2}{3} \left[8 - 1\right]$

$A = \frac{2}{3} [7]$

$A = \frac{14}{3}$

The area of the region is

\frac{14}{3}

square units.

Question 2

Find the area of the region bounded by $y^2 = 9x$ , $x = 2$ , $x = 4$ , and the x-axis in the first quadrant.

Solution: The region is bounded by the curve

y^2 = 9x

, the vertical lines

x = 2

and

x = 4

, and the x-axis. We are considering the first quadrant, so

y \ge 0

From $y^2 = 9x$ , we get $y = \sqrt{9x} = 3\sqrt{x}$ (taking the positive root for the first quadrant).

The area

A

is found by integrating

y

with respect to

x

from

x = 2

x = 4

$A = \int_{2}^{4} y \, dx$

Substitute

y = 3\sqrt{x}

$A = \int_{2}^{4} 3\sqrt{x} \, dx$

Evaluate the integral:

$A = 3 \left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{2}^{4}$

$A = 3 \cdot \frac{2}{3} \left[x^{\frac{3}{2}}\right]_{2}^{4}$

$A = 2 \left[x^{\frac{3}{2}}\right]_{2}^{4}$

Substitute the limits of integration:

$A = 2 \left[\left(4\right)^{\frac{3}{2}} - \left(2\right)^{\frac{3}{2}}\right]$

$A = 2 \left[8 - 2\sqrt{2}\right]$

$A = 16 - 4\sqrt{2}$

The area of the region is

16 - 4\sqrt{2}

square units.

Question 3

Find the area of the region bounded by $x^2 = 4y$ , $y = 2$ , $y = 4$ , and the y-axis in the first quadrant.

Solution: The region is bounded by the curve

x^2 = 4y

, the horizontal lines

y = 2

and

y = 4

, and the y-axis. We are considering the first quadrant, so

x \ge 0

From $x^2 = 4y$ , we get $x = \sqrt{4y} = 2\sqrt{y}$ (taking the positive root for the first quadrant).

To find the area bounded by a curve

x = g(y)

and the y-axis, we integrate

x

with respect to

y

between the given y-limits.

The area $A$ is calculated as:

$A = \int_{2}^{4} x \, dy$

Substitute

x = 2\sqrt{y}

$A = \int_{2}^{4} 2\sqrt{y} \, dy$

Evaluate the integral:

$A = 2 \left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{2}^{4}$

$A = 2 \cdot \frac{2}{3} \left[y^{\frac{3}{2}}\right]_{2}^{4}$

$A = \frac{4}{3} \left[y^{\frac{3}{2}}\right]_{2}^{4}$

Substitute the limits of integration:

$A = \frac{4}{3} \left[\left(4\right)^{\frac{3}{2}} - \left(2\right)^{\frac{3}{2}}\right]$

$A = \frac{4}{3} \left[8 - 2\sqrt{2}\right]$

$A = \frac{32 - 8\sqrt{2}}{3}$

The area of the region is

\frac{32 - 8\sqrt{2}}{3}

square units.

Common mistakes

Incorrectly identifying the limits of integration.
Errors in setting up the integrand based on the curve equation.
Mistakes during the evaluation of the definite integral.
Confusing x-axis boundaries with y-axis boundaries.

Revision tips

Sketch the region for each problem to visualize the boundaries.
Ensure the limits of integration are correctly determined from the given lines or intersection points.
Practice rewriting the curve equation to express y in terms of x (or vice versa) as needed for integration.
Double-check the integration and substitution steps during evaluation.

Practice MCQs

Q1. What is the general approach to find the area bounded by a curve y = f(x) and the x-axis between x = a and x = b?

Q2. For the curve $y^2 = x$ , what is y expressed in terms of x for the first quadrant?

Q3. When finding the area bounded by $x^2 = 4y and the y-axis$ , which variable should be integrated with respect to?

Q4. The integral $ $\int$ _{1}^{4} $\sqrt{x}$ dx$ represents the area bounded by which curve and lines?

Frequently asked questions

What is the main concept covered in Chapter 8 of CBSE Class 12 Maths NCERT Solutions?

Chapter 8 focuses on the Applications of Integrals, specifically using definite integrals to calculate the area of regions bounded by curves, lines, and axes.

How do these NCERT Solutions help students?

These solutions provide clear, step-by-step explanations for solving problems related to finding areas, helping students understand the methods and practice for exams.

What types of curves are typically involved in these area problems?

The problems usually involve simple curves like parabolas (e.g., y^2 = x, x^2 = 4y) and straight lines, often focusing on areas in the first quadrant.

Is sketching the region important for solving these problems?

Yes, sketching the region bounded by the given curves and lines is crucial for correctly identifying the limits of integration and setting up the integral.

How is the area calculated when the region is bounded by y^2 = x and the x-axis?

The area is found by integrating y = $\sqrt{x}$ with respect to x between the given x-limits, as y = $\sqrt{x}$ represents the upper boundary in the first quadrant.

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CBSE Class 12 Mathematics Chapter 8: Applications of Integrals NCERT Solutions

Quick info

Chapter summary

Learning outcomes

Topics covered

Paper topics

Important topics

PDF preview

Questions and Solutions

Question 1

Question 2

Question 3

Common mistakes

Revision tips

Practice MCQs

Frequently asked questions

Important links