NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

Imagine you have a curved shape like the edge of a hill or a wave, and you want to find out how much space it takes up between the curve and the ground (x-axis). That space under the curve is called the area, and to calculate this area, we use integration. Integrals are the functions that satisfy a given differential equation for finding the area of a curvy region y = f(x), the x-axis, and the line x = a and x = b (b > a). Therefore, finding the integral of a function concerning x means finding the area to the X-axis from the curve. The integral is also called anti-derivative as it is the reverse process of differentiation.

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This Story also Contains

1. Application Of Integrals Class 12 Maths Chapter 8 - PDF Free Download
2. Application Of Integrals Class 12 NCERT Solutions - Important Formulae
3. NCERT Application Of Integrals Class 12 Questions And Answers (Exercise)
4. NCERT solutions for class 12 maths - Chapter-wise
5. Importance of NCERT Solutions Class 12 Chapter 8:
6. NCERT solutions for class 12 subject-wise
7. NCERT Solutions Class Wise
8. NCERT Books and NCERT Syllabus

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Interested students can find all NCERT Solutions for Class 12 Maths in one place. In the applications of integrals class 12 ncert solutions, questions from all these topics are covered. If you are interested check all NCERT solutions from classes 6 to 12 in a single place, which will help you to learn CBSE maths and science. Students can additionally refer to NCERT Exemplar Solutions for Class 12 Maths Chapter 8 Applications of Integrals for better practice and understanding of the topic.

Application Of Integrals Class 12 Maths Chapter 8 - PDF Free Download

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Application Of Integrals Class 12 NCERT Solutions - Important Formulae

1. Area Enclosed by a Curve and Lines:

The area enclosed by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ (where $b>a$ ) is given by the formula:

Area $={\int }_a^{b}ydx={\int }_a^{b}f(x)dx$

2. Area Bounded by Curve and Horizontal Lines:

The area of the region bounded by the curve $x=\varphi (y)$ as its $y$-axis and the lines $y=c$ and $y=d$ is given by the formula:

Area $={\int }_c^{d}xdy={\int }_c^d\varphi (y)dy$

3. Area Between Two Curves and Vertical Lines:

The area enclosed between two given curves $y=f(x)$ and $y=g(x)$, and the lines $x=a$ and $x=b$, is given by the formula:

Area $={\int }_a^b[f(x)-g(x)]dx($ Where $f(x)\ge g(x)$ in $[a,b])$

4. Area Between Curves with Different Intervals:

If $f(x)\ge g(x)$ in $[a,c]$ and $f(x)\le g(x)$ in $[c,b]$, where $a<c<b$, then the resultant area between the curves is given as:

Area $={\int }_a^c[f(x)-g(x)]dx+{\int }_c^b[g(x)-f(x)]dx$

NCERT Application Of Integrals Class 12 Questions And Answers (Exercise)

Question 1: Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Answer:

The area bounded by the ellipse : $\frac{x^2}{16}+\frac{y^2}{9}=1.$

The area will be 4 times the area of EAB.

Therefore, $AreaofEAB={\int }_0^{4}ydx$

$={\int }_0^{4}3\sqrt{1-\frac{x^2}{16}}dx$

$=\frac{3}{4}{\int }_0^4\sqrt{16-x^2}dx$

$=\frac{3}{4}{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_0^4$

$=\frac{3}{4}[2\sqrt{16-16}+8{\sin }^{-1}(1)-0-8{\sin }^{-1}(0)]$

$=\frac{3}{4}\left[\frac{8\pi }{2}\right]$

$=\frac{3}{4}\left[4\pi \right]=3\pi$

Therefore the area bounded by the ellipse will be $=4\times 3\pi =12\pi units.$

Que,stion 2: Find the area of the region bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Answer:

The area bounded by the ellipse : $\frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore, $AreaofEAB={\int }_0^{2}ydx$

$={\int }_0^{2}3\sqrt{1-\frac{x^2}{4}}dx$

$=\frac{3}{2}{\int }_0^2\sqrt{4-x^2}dx$

$=\frac{3}{2}{[\frac{x}{2}\sqrt{4}-x^2+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^2$

$=\frac{3}{2}\left[\frac{2\pi }{2}\right]$

$=\frac{3\pi }{2}$

Therefore the area bounded by the ellipse will be $=4\times \frac{3\pi }{2}=6\pi units,.$

Question 3: Choose the correct answer in the following

The area lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

$(A)\pi$ $(B)\frac{\pi }{2}$ $(C)\frac{\pi }{3}$ $(D)\frac{\pi }{4}$

Answer:

The correct answer is A
The area bounded by circle C(0,0,4) and the line x=2 is

The required area = area of OAB
${\int }_0^{2}ydx={\int }_0^2\sqrt{4-x^2}dx$
$$\begin{gathered} =[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}{]}_0^2 \\ =2(\pi /2) \\ =\pi \end{gathered}$$

Question 4: Choose the correct answer in the following.

Area of the region bounded by the curve $y^2=4x$ , $y$ -axis and the line $y=3$ is

(A) $2$ (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$

Answer:

The area bounded by the curve $y^2=4x$ and y =3

The required area = OAB =
$$\begin{gathered} {\int }_0^{3}xdy \\ ={\int }_0^3\frac{y^2}{4}dy \\ =\frac{1}{4}.[\frac{y^3}{3}{]}_0^3 \\ =\frac{9}{4} \end{gathered}$$

Question 1: Find the area under the given curves and given lines:

(i) $y=x^2,x=1,x=2$ and $x$ -axis

Answer:

The area bounded by the curve $y=x^2,x=1,x=2$ and $x$ -axis

The area of the required region = area of ABCD
$$\begin{gathered} ={\int }_1^{2}ydx \\ ={\int }_1^2{x}^{2}dx \\ =[\frac{x^3}{3}{]}_1^2 \\ =\frac{7}{3} \end{gathered}$$
Hence the area of the shaded region is 7/3 units.

Question:1 Find the area under the given curves and given lines:

(ii) $y=x^4,x=1,x=5$ and $x$ -axis

Answer:

The area bounded by the curev $y=x^4,x=1,x=5$ and $x$ -axis

The area of the required region = area of ABCD
$$\begin{gathered} ={\int }_1^{5}ydx \\ ={\int }_1^2{x}^{4}dx \\ =[\frac{x^5}{5}{]}_1^2 \\ =625-\frac{1}{5} \\ =624.8 \end{gathered}$$
Hence the area of the shaded region is 624.8 units.

Question 2: Sketch the graph of $y=|x+3|$ and evaluate ${\int }_{-6}^0|x+3|dx.$

Answer:

y=|x+3|

The given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

The integral to be evaluated is

${\int }_{-6}^0|x+3|dx$

$={\int }_{-6}^{-3}(-x-3)dx+{\int }_{-3}^0(x+3)dx$

$=[-\frac{x^2}{2}-3x{]}_{-6}^{-3}+[\frac{x^2}{2}+3x{]}_{-3}^0$

$$\begin{gathered} =(-\frac{9}{2}+9)-(-18+18)+0-(\frac{9}{2}-9) \\ =9 \end{gathered}$$

Question 3: Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2\pi$.

Answer:

The graph of y=sinx is as follows

We need to find the area of the shaded region.

ar(OAB)+ar(BCD)

=2ar(OAB)

$$\begin{gathered} =2\times {\int }_0^{\pi }sinxdx \\ =2\times [-cosx{]}_0^{\pi } \\ =2\times [-(-1)-(-1)] \\ =4 \end{gathered}$$

The bounded area is 4 units.

Question 4: Choose the correct answer.

Area bounded by the curve $y=x^3$ , the $x$ -axis and the ordinates $x=-2$ and $x=1$ is

(A) $-9$ (B) $\frac{-15}{4}$ (C) $\frac{15}{4}$ (D) $\frac{17}{4}$

Answer:

Hence, the required area

$={\int }_{-2}^{1}ydx$

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

$={\left[\frac{x^4}{4}\right]}_{-2}^0+{\left[\frac{x^4}{4}\right]}_0^1$

$=[0-\frac{(-2{)}^4}{4}]+[\frac{1}{4}-0]$

$=-4+\frac{1}{4}=\frac{-15}{4}$

Therefore, the correct answer is B.

Question 5: Choose the correct answer.

The area bounded by the curve $y=x|x|$ , $x$ -axis and the ordinates $x=-1$ and $x=1$ is given by

(A) $0$ (B) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $\frac{4}{3}$

[ Hint : $y=x^2$ if $x>0$ and $y=-x^2$ if $x<0$ . ]

Answer:

The required area is

$$\begin{gathered} 2{\int }_0^1{x}^{2}dx \\ =2{\left[\frac{x^3}{3}\right]}_0^1 \\ =\frac{2}{3}units \end{gathered}$$

If you are looking for exercise solutions for the chapter application of integrals class 12 then these are listed below.

• Applications Of Integrals Class 12 Exercise 8.1
• Ap,plication Of Integrals Class 12 Miscellaneous Exercise

Also read,

• NCERT Exemplar Class 12 Chemistry Solutions
• NCERT Exemplar Class 12 Mathematics Solutions
• NCERT Exemplar Class 12 Biology Solutions
• NCERT Exemplar Class 12 Physics Solutions

NCERT solutions for class 12 maths - Chapter-wise

Importance of NCERT Solutions Class 12 Chapter 8:

• These solutions are very easy to understand as they are explained and prepared in a detailed manner.
• Scoring good marks in the 12th board exam is now a reality with the help of these solutions.
• NCERT Solutions for Class 12 Maths Chapter 8 PDF download provides in-depth knowledge of the concepts as well as different ways to solve the problems.
• These solutions prepared by experts who know how to answer to score well in the board exam.
• You will also get some short tricks to check whether the answer is correct or not.

NCERT solutions for class 12 subject-wise

Students can check the following links for more in-depth learning.

• NCERT solutions for class 12 mathematics
• NCERT Solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions Class Wise

Students can check the following links for more in-depth learning.

• NCERT solutions for class 12
• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Students can check the following links for more in-depth learning.

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12

Happy Reading !!!