Application of Integrals

Application of Integrals in Mathematics

Integration is the reverse process of differentiation. In integration, we determine the function whose derivative is given. Class 12 application of integrals formulas rely on this principle to solve real-life and mathematical problems.

Integrals are based on a limiting process that approximates the area of a curvilinear region by dividing it into thin vertical slabs and summing them up.

For example: $\frac{d}{dx} (\sin x) = \cos x$
$\frac{d}{dx} (x^2) = 2x$
$\frac{d}{dx} (e^x) = e^x$

In these examples, $\cos(x)$ is the derivative of $\sin(x)$. So, $\sin(x)$ is an antiderivative (or integral) of $\cos(x)$. Similarly, $x^2$ and $e^x$ are antiderivatives (or integrals) of $2x$ and $e^x$, respectively.

Role of the Constant of Integration

The derivative of a constant ($C$) is zero. Therefore, we can also write:

$\frac{d}{dx} (\sin x + C) = \cos x$
$\frac{d}{dx} (x^2 + C) = 2x$
$\frac{d}{dx} (e^x + C) = e^x$

This shows that the antiderivatives (or integrals) of a function are not unique. There are infinitely many integrals for each function, obtained by choosing $C$ arbitrarily from real numbers. This principle forms the foundation of many applications of integrals, class 12 solutions and exercises.

Formulae for Integration

This section provides all essential formulas for indefinite and definite integrals, covering polynomials, trigonometric, exponential, logarithmic, and inverse functions. Mastering these formulas is key to solving the application of integrals class 12 problems efficiently.

Formulas for Indefinite Integrals

This section lists all key formulas for indefinite integrals, including polynomials, trigonometric, exponential, and logarithmic functions, essential for solving class 12 integrals problems.

Formulas for Definite Integrals

Here, you’ll find important formulas for definite integrals to calculate areas, volumes, and accumulated quantities in class 12 application of integrals problems.

Key Properties of Definite Integrals

1. Change of variable: $\int_a^b f(x) ,dx = \int_a^b f(t) ,dt$

2. Reversing limits: $\int_a^b f(x) ,dx = -\int_b^a f(x) ,dx$

3. Integral over same limits: $\int_a^a f(x) ,dx = 0$

4. Breaking into parts: $\int_a^b f(x) ,dx = \int_a^c f(x) ,dx + \int_c^b f(x) ,dx$

5. Even function: $\int_{-a}^a f(x) ,dx = 2 \int_0^a f(x) ,dx, \quad f(-x) = f(x)$

6. Odd function: $\int_{-a}^a f(x) ,dx = 0, \quad f(-x) = -f(x)$

Application of Integrals Class 12

The application of Integrals class 12 notes includes - the area along the X-axis and Y-axis, the area of a piecewise function and the area bounded by two curves.

Area along the X-axis

If the function $f(x) ≥ 0 ∀ x ∈ [a, b]$ then $\int_a^{\infty} f(x) d x$ represents the area bounded by $y = f(x), x-$axis and lines $x = a$ and $x = b$.

If the function $f(x) ≤ 0 ∀ x ∈ [a, b]$, then the area by bounded $4y = f(x)$, x-axis and lines $4x = a$ and $x = b$ is $\int_a^b f(x) d x$.

Area along the Y-axis

The area by bounded $x = g(y)$ [with $g(y)>0$], $y$-axis and the lines $y = a$ and $y = b$ is $\int_a^b x d y=\int_a^b g(y) d y$

Area of Piecewise Function

If the graph of the function $f(x)$ is of the following form, then

then $\int_a^b f(x) d x$ will equal $A_1-A_2+A_3-A_4$ and not $A_1+A_2+A_3+A_4$.

If we need to evaluate $A_1+A_2+A_3+A_1$ (the magnitude of the bounded area) we will have to calculate

$ \underbrace{\int_a^x f(x) d x}_{\mathrm{A}_1}+\underbrace{\left|\int_x^y f(x) d x\right|}_{\mathrm{A}_2}+\underbrace{\int_5^z f(x) d x}_{\mathrm{A}_3}+\underbrace{\left|\int_z^b f(x) d x\right|}_{\mathrm{A}_4} $

The area bounded by the curve when the curve intersects $X$-axis

The graph $y=f(x) \forall x \in[a, b]$ intersects $x-a x i s$ at $x=c$.

If the function $f(x) \geq 0 \forall x \in[a, c]$ and $f(x) \leq 0 \forall x \in[c, b]$ then area bounded by curve and $x$-axis, between lines $x=a$ and $x=b$ is

$ \int_a^b|f(x)| d x=\int_a^c f(x) d x-\int_c^b f(x) d x$

Area Bounded by Two Curves

Area bounded by the curves $y=f(x), y=g(x) $ and the lines $ x = a$ and $x = b$, and it is given that $f(x) ≤ g(x). $

From the figure, it is clear that,

Area of the shaded region = Area of the region $ABEF$ - Area of the region $ABCD$

$\int_a^b g(x) d x-\int_a^b f(x) d x=\int_a^b(\underbrace{g(x)}_{\begin{array}{c}\text { upper } \\ \text { curve }\end{array}}-\underbrace{f(x)}_{\begin{array}{c}\text { lower } \\ \text { curve }\end{array}}) d x$

Area Bounded by Curves When Intersecting at More Than One Point

Area bounded by the curves $y = f(x), y = g(x)$ which intersect each other in the interval $[a, b]$

First find the point of intersection of these curves $y = f(x)$ and $y = g(x)$ by solving the equation $f(x) = g(x)$, let the point of intersection be $x = c $

$=\int_a^c\{f(x)-g(x)\} d x+\int_c^b\{g(x)-f(x)\} d x$

When two curves intersects more than one point

Area bounded by the curves $y=f(x), y=g(x)$ which intersect each other at three points at $x = a, x = b$ and $x = c. $

To find the point of intersection, solve $f(x) = g(x). $

For $x ∈ (a, c), f(x) > g(x)$ and for $x ∈ (c, b),g(x) > f(x).$

Area bounded by curves,

$\begin{aligned} A & =\int_a^b|f(x)-g(x)| d x \\ & =\int_a^c(f(x)-g(x)) d x+\int^b(g(x)-f(x)) d x\end{aligned}$

Now, let us summarise and recall the application of the integrals class 12 formulas.

Application of the Integrals Formula

Application of integrals class 12 formulas includes the formulas for the area along the X-axis and Y-axis, area of a piecewise function, and are bounded by two curves.

Area along the X-axis

If the function $f(x) ≥ 0 ∀ x ∈ [a, b]$ then $\int_a^{\infty} f(x) d x$ represents the area bounded by $y = f(x), x-$axis and lines $x = a$ and $x = b$.

If the function $f(x) ≤ 0 ∀ x ∈ [a, b]$, then the area by bounded $4y = f(x)$, x-axis and lines $4x = a$ and $x = b$ is $\int_a^b f(x) d x$.

Area along the Y-axis

The area by bounded $x = g(y)$ [with $g(y)>0$], $y$-axis and the lines $y = a$ and $y = b$ is $\int_a^b x d y=\int_a^b g(y) d y$

Area of Piecewise Function

If the function $f(x) \geq 0 \forall x \in[a, c]$ and $f(x) \leq 0 \forall x \in[c, b]$ then area bounded by curve and $x$-axis, between lines $x=a$ and $x=b$ is

$ \int_a^b|f(x)| d x=\int_a^c f(x) d x-\int_c^b f(x) d x$

Area Bounded by Two Curves

Area bounded by the curves $y=f(x), y=g(x) $ and the lines $ x = a$ and $x = b$, and it is given that $f(x) ≤ g(x). $

Area bounded by the curves = $\int_a^b g(x) d x-\int_a^b f(x) d x=\int_a^b(\underbrace{g(x)}_{\begin{array}{c}\text { upper } \\ \text { curve }\end{array}}-\underbrace{f(x)}_{\begin{array}{c}\text { lower } \\ \text { curve }\end{array}}) d x$

Application of Integrals in Mathematics: Solved Previous Year Questions

Question 1:

For the given curve, $y=a \sin t ; x=a \cos t+\frac{a}{2} \ln \tan ^{2} \frac{t}{2}$ , the area bounded by the curve is:

Solution:

Required Area =

$
\begin{aligned}
& =4 \int_0^{\pi / 2} \mathrm{y} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{dt} \\
= & 4 \int_0^{\pi / 2}(\mathrm{a} \sin \mathrm{t})\left(\frac{\mathrm{a}}{\sin t} \cos ^2 \mathrm{t}\right) \mathrm{dt} \\
= & 4 \mathrm{a}^2 \int_0^{\pi / 2} \cos ^2 \mathrm{tdt}=\pi \mathrm{a}^2
\end{aligned}
$

Hence, the correct answer is $\pi a^2$.

Question 2:

The area of region which satisfy $x^4+x^3 y^3 \leq x y+y^4$ and $x^2+y^2 \leq 2$ is:

Solution:

$
\begin{aligned}
& y^4+x y-x^4-x^3 y^3 \geq 0 \\
\rightarrow & \left(y^3+x\right)\left(y-x^3\right) \geq 0
\end{aligned}
$
Thus above curves partition the circle into 4 regions of equal area, and inequality is satisfied by two regions. So, the area is half the area of a circle.

Hence, the correct answer is the $\pi$.

Question 3:

The area of the region bounded by the curve x = 2y + 3 and the y-axis. y = 1 and y = –1 is:

Solution:

The question mentions the curve $\mathrm{x}=2 \mathrm{y}+3$ and the lines on y axis $\mathrm{y}=1$ and $\mathrm{y}=-1$

Required area

$
\begin{aligned}
& =\int_{-1}^1(2 y+3) d x \\
& =\left[y^2+3 y\right]_{-1}^1 \\
& =(1+3-1+3) \\
& =6 \text { sq.units }
\end{aligned}
$
Hence, the correct answer is 6 sq units.

Question 4:

The area of the region bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is:

Solution:

Ellipse has a symmetry with $x$ axis and $y$ axis
The required area can be calculated as

$
\begin{aligned}
& =4 \int_0^5 \frac{4}{5}\left(\sqrt{25-x^2}\right) d x \\
& {\left[\int \sqrt{a^2-x^2} d x=\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\
& =\frac{16}{5} \int_0^5\left(\sqrt{5^2-x^2}\right) d x \\
& =\frac{16}{5}\left[\frac{x \sqrt{5^2-x^2}}{2}+\frac{5^2}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right]_0^5 \\
& =\frac{16}{5}\left(0-\frac{25 \pi}{4}-0-0\right) \\
& =20 \pi \text { sq.units }
\end{aligned}
$

Hence, the correct answer is 20π sq units.

Question 5:

Area of the region bounded by the curve $y = cos x$ between $x = 0$ and $x = \pi$ is:

Solution:

The questions mention curve $y = cos x$ and $x = 0$ and $x = \pi$

$\begin{aligned}
&\text { Required area }\\
&\begin{aligned}
& =\int_0^\pi|\cos x| d x \\
= & 2 \int_0^{\frac{\pi}{2}} \cos x d x \\
= & 2[\sin x]_0^{\frac{\pi}{2}} \\
= & 2(1-0) \\
= & 2 \text { sq.units }
\end{aligned}
\end{aligned}$

Hence, the correct answer is 2 sq units.

List of Topics related to applications of Integrals according to NCERT/JEE Mains

This section lists all essential topics on integrals from NCERT that are frequently asked in JEE Main, helping you focus on important areas.

Application of Integrals in Different Exams

The chapter "Application of Integrals" focuses on using integration techniques to find the areas of regions bounded by curves and straight lines. It is an important part of Class 12 Mathematics and is frequently tested in both board and competitive examinations. Questions from this chapter assess a student’s ability to analyse graphs, set up proper limits of integration, and apply definite integrals accurately. Regular practice of NCERT problems and graph-based questions helps students master this chapter and achieve a good score in exams.

Important Books and Resources for Class 12 Integrals

Here, you’ll find the best books and reference materials to strengthen your understanding of integrals and practice effectively for exams.

NCERT Resources for Application of Integrals

This section highlights NCERT textbooks and reference material that clearly explain integrals, forming the foundation for solving problems.

• NCERT Maths Notes for Class 12th Chapter 8 Applications of Integrals

• NCERT Maths Solutions for Class 12th Chapter 8 Applications of Integrals

• NCERT Maths Exemplar Solutions for Class 12th Chapter 8 Applications of Integrals

NCERT Subjectwise Resources

Explore subject-wise NCERT resources that cover notes, solutions and exemplar problems, supporting structured and focused learning.

Practice Questions based on Applications of Integrals

This section provides important practice questions and exercises based on integrals to improve problem-solving skills and exam readiness.

Conclusion

The chapter Application of Integrals plays a crucial role in understanding how integration is used to solve real-life problems involving continuously varying quantities. By learning to calculate areas bounded by curves and straight lines, students develop strong graphical and analytical skills. This chapter strengthens the practical application of calculus and is an important scoring area in Class 12 Mathematics. With regular practice of NCERT problems, proper graph analysis, and correct use of limits of integration, students can master this chapter and perform confidently in board as well as competitive examinations.