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The branch of calculus where we examine integrals and their properties is known as **integral calculus**. Integration is a crucial concept because it is the inverse of differentiation. The fundamental theorem of calculus connects integral and differential calculus.

In this article, we will go over integral calculus in detail, including why it's employed, its types, formulas, examples, and applications. Also, for a better understanding and knowledge, we will go over some of the solved problems and previous year's questions.

Indefinite integral

Properties of indefinite integral

Methods of integration

Integration by substitution

Integration using trigonometric functions

Integration by partial fractions

Integration by parts

Integrals of some particular function

Integral of some special types

Definite integral and its properties

The fundamental theorem of calculus

Evaluations of definite integral by substitution

The values of the function discovered through the integration process are known as integrals. Integration is the process of obtaining f(x) from f'(x). Integrals assign numbers to functions in a way that represents displacement and motion problems, area and volume problems, and other problems that arise when little data is combined. We can find the function f by using the derivative f' of the function f and that function f is referred to as antiderivative or integral of f' in this case.

Integral calculus is used to solve problems of the following types given below:

A. The first type of problem is to find the function if its derivative is given.

B. Also, the problem is related to finding the area bounded by the graph of a function under given conditions. Further, the Integral calculus is divided into two types.

Definite Integrals (in which the value of the integrals are definite)

Indefinite Integrals (in which the value of the integral is indefinite with an arbitrary constant, C)

The integrals which do not have a pre-existing value of limits, which makes the final value of integral indefinite, I.e, ∫g'(x)dx = g(x) + c. Indefinite integrals are part of the family parallel curves.

Definite integral is one that has a pre-existing value of limits, which makes the final value of the integral, definite. if a function of the curve is f(x) , then $\int\limits_a^b f(x) dx = f(b) - f(a)$

Image: Definite Integral

Integrals are defined as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, at X-axis where the ordinates are x = a and x =b, and b>a. Let x be a given point in $[$a,b$]$. Then $\int\limits_a^b f(x) dx$ represents the area function. This concept of area function leads to the fundamental theorems of integral calculus.

First Fundamental Theorem of Integral Calculus

Second Fundamental Theorem of Integral Calculus

A(x) = $\int\limits_a^b f(x) dx$ for all x ≥ a, where the function is continuous on $[$a,b$]$. Then A'(x) = f(x) for all x ϵ $[$a,b$]$

If f is continuous function of x defined on the closed interval $[$a,b$]$ and F is another function such that d/dx F(x) = f(x) for all x in the domain of f, then $\int\limits_a^b f(x) dx$ = f(b) -f(a). This is known as the definite integral of f over the range $[$a,b$]$, Where a is the lower limit and b is the upper limit.

Let us go through some of the properties of the indefinite integrals:

The derivative of an integral is the integrand of itself, ∫ f(x) dx = f(x) +C.

Since two indefinite integrals with the same derivative produce the same family of curves, they are equivalent, ∫ $[$ f(x) dx - g(x) dx$]$ = 0.

The sum or the difference of a finite number of functions is equal to the sum or difference of the individual functions' integrals, ∫ $[$ f(x) dx+g(x) dx$]$ = ∫ f(x) dx + ∫ g(x) dx.

The constant is taken outside the integral sign, ∫ k f(x) dx = k ∫ f(x) dx, where k ∈ R.

The previous two properties are combined to get the form: ∫ $[k_1 f_1(x)$ + $k_2 f_2(x)$ +... $k_n f_n(x)]\; dx$ = $k_1 \lim f_1(x)\;dx$ + $k_2 \lim f_2(x)\;dx$+ ... $k_n \lim f_n(x)\;dx$

We have different methods to find the integral of a given function in integral calculus. Some of the most commonly used methods of integration are:

Integration by Parts

Integration using Substitution

Integration by Partial Fractions

Integral Calculus is mainly used for the two purposes mentioned below:

To calculate f from f’: If a function f is differentiable in the interval of consideration, then f’ is defined. We've already seen how to calculate a function's derivatives in differential calculus, and we can "undo" that with the help of integral calculus.

To find the area under a curve.

Some of the important applications of the integral calculus are - Integration is applied to find:

The area between two curves

Centre of mass

Kinetic energy

Surface area

Work

Distance, velocity and acceleration

The average value of a function

Volume

Example 1: Find the integral of the function f(x) = $\sqrt{x}$.

Solution:

Given,

f(x) = $\sqrt{x}$

∫f(x) dx = $\int \sqrt{x} dx$

$\begin{array}{l}\int \sqrt{x}\ dx = \int x^{\frac{1}{2}}\ dx\end{array} $

We know that, $\begin{array}{l}\int x^{n}\ dx = \frac{x^{n+1}}{n+1}+C\end{array} $

Now, $\begin{array}{l}\int \sqrt{x}\ dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\\= \frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}+C\\=\frac{2}{3}x^{\frac{3}{2}}+C\end{array} $

Example 2: Find the integral of cos2n with respect to n.

Solution:

Let f(n) = cos2n

we know that 2 cos2A = cos 2A + 1

So, f(n) = (1/2)(cos 2n + 1)

Let us find the integral of f(n).

∫f(n) dn = ∫(1/2)(cos 2n + 1) dn

= (1/2) ∫(cos 2n + 1) dn

= (1/2) ∫cos 2n dn + (1/2)∫1 dn

= (1/2) (sin 2n/2) + (1/2) n + C

= (sin 2n/4) + (n/4) + C

=(1/4)$[$sin 2n + n$]$ + C

Question 1: ∫{$[$sin8x − cos8x$]$ / $[$1 − 2 sin2x cos2x$]$} dx = _________.

Solution:

∫{$[$sin8x − cos8x$]$ / $[$1 − 2 sin2x cos2x$]$} dx

= ∫{$[$(sin4x + cos4x) * (sin4x − cos4x)$]$ / $[$(sin2x + cos2x)2 − 2 sin2x cos2x$]$} dx

= ∫(sin4x – cos4x) dx

= ∫$[$sin2x + cos2x$] \times [$sin2x – cos2x$]$ dx

= ∫(sin2x - cos2x) dx

= ∫−cos2x dx

= $[$−sin2x/2$]$ + c

Question 2: ∫x2dx / (a + bx)2 = ___________.

Solution:

$ \int \frac{x^{2}}{(a+b x)^{2}} d x $

Let $ a+b x=t \Rightarrow b\; d x=d t $

$ s_{0}, \frac{1}{b} \int \left(\frac{t-a}{b t}\right)^{2} d t=\frac{1}{b^{3}} \int\left(1-\frac{a}{t}\right)^{2} d t $

$ =\frac{1}{b^{3}} \int\left(1+\frac{a^{2}}{t^{2}}-\frac{2 a}{t}\right) d t $

$=\frac{1}{b^3}\left [ \int dt +a^2\int\frac{dt}{t^2}-2a\int \frac{dt}{t}\right ]$

$ =\frac{1}{b^{3}}\left[ t+a^{2} \left( -\frac{1}{t} \right) -2 a\; ln\; t \right]+c $

$ =\frac{a+b x}{b^{3}}-\frac{a^{2}}{b^{3}(a+b x)}-\frac{2 a}{b^{3}} ln (a+bx)+c $

$ =\frac{x}{b^{2}}-\frac{a^{2}}{b^{3}(a+b x)}-\frac{2 a}{b^{3}} ln (a+bx)+k $

Question 3: $\int \left [ {x^5}/{\sqrt{1+x^3}} \right ] dx$ = ________.

Solution:

Put 1 + x3 = t2

⇒ 3x2 dx = 2tdt and x3 = t2 − 1

So, ∫$x^5 /\sqrt{1+x^3} \; dx$ = ∫{$[x^2$ $\times$ $x^3]$ / $\sqrt{1+x^3} \; dx$

= $\left [ \dfrac{2}{3} \right ]$ ∫{$[$(t2 − 1) $\times$ t$]$ dt / $[$t$]$}

= $\left [ \dfrac{2}{3} \right ]$ ∫(t2 − 1) dt

= $\left [ \dfrac{2}{3} \right ]$ $[$(t3 / 3) − t$]$ + c

= $\left [ \dfrac{2}{3} \right ]$ $[${(1 + x3)3/2 / 3} − {(1 + x3)½}$]$+ c

1. Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x $\in$0, 1], p(0) = 1 and p(1) = 41. Then $\int_{0}^{1}$ p(x) dx equals:

(A) $\sqrt{41}$

(B) 21

(C) 41

(D) 42

2. The value of $\int_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$ is:

(A) $\frac{\pi}{2} \log 2$

(B) $\log 2$

(C) $\pi \log 2$

(D) $\frac{\pi}{8} \log 2$

Answers: 1. (B), 2. (C)

Integral calculus is an extremely important concept to learn. Integral calculus is a branch of calculus that includes differential calculus and is utilised in physics. Aspirants preparing for the JEE Main exam can use this article to learn the fundamentals of integrals such as definition, types of integrals, and properties. They can also look at the solved examples and previous year's problems to gain a better understanding and knowledge.

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It is crucial for the the engineering aspirants to know and download the JEE Advanced 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Advanced 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Advanced 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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IIT Bombay is responsible for the release of the JEE Advanced 2023 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Advanced 2023 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Advanced qualifying marks for 2021 ranged from 17.50%, while for OBC/SC/ST categories, they ranged from 15.75% for OBC, 8.75% for SC and 8.75% for ST category.

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FAQ

**1. What is Integral Calculus Used For?**

We use definite integrals to determine the area under the curve or between the curves given by functions, then we use formulae and procedures to discover their indefinite integrals, and then we utilise limits to find their difference of integrals. For computing the volumes of 3-d solids, we employ definite integrals. We can find the distance given the velocity since the distance is the integral of velocity.

**2. What is the contribution of chapter JEE main mathematics sets, relations and functions?**

Integral Calculus is a crucial topic to learn; the complexity of questions from this topic is medium, and they will be simple to solve once you have mastered the concept. Every year, you will encounter 1 - 3 questions in JEE Main and other tests, directly (since the chapter weight in JEE Main is only 9%), but also indirectly (because the application of this chapter is involved in physics and area finding problems).

**3. How do you use Integrals using Trigonometry?**

To find the integrals using trigonometry, use the trigonometric identities to simplify the function into an integral function, then use the formulae and integration techniques to determine the integrals.

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