In Mathematics, Integral Calculus is one of the very important topics to understand, the difficulty of question from this topic is medium and will be easy to solve if you are through with the concept. Every year you will get at max 1  3 questions in JEE Main and other exams, directly (as chapter weight in jee main is only 9%) but indirectly, the application of this chapter will be involved in physics and area finding problems. This chapter will be helpful in mechanics to solve the problem which involves integration. It will be a new chapter for student and you will find it difficult to learn initially but with time once you understand the basic concept and do the proper amount of practice you will find it handy. This chapter is highly calculative, so you need to do proper practice, you should solve an ample amount of questions and make sure you are good with calculations. Overall this chapter will be larger than other chapters as it has three topics combined.
Integral Calculus is important due to various reallife cases and the handy tool it provides for various reallife application. It is used in finance to price bonds and options, credit card companies use integral calculus to set the due on credit cards. In chemistry, the rate of reaction is determined by using the integral calculus. The area finding the application of integral calculus is used by architectengineer to determine the need for material for the construction. Graphic design artists use integral calculus to determine how different threedimensional model will behave when subjected to rapidly changing conditions.
If you understand this chapter well you will realize how good tools this chapter provide to you. You will love to solve questions based on this chapter.
Important topics :
Introduction, integration as the inverse function of differentiation
Indefinite integral and properties of indefinite integral
Comparison between definite and indefinite integration
Methods of integration, Integration by substitution, integration using trigonometric identities, integration by partial fractions, integration by parts, etc
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
Introduction: Differential Calculus is centered on the concept of the derivative. The original motivation for the derivative was the problem of defining tangent lines to the graphs of functions and calculating the slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions.
a) The problem of finding a function whenever its derivative is given,
b) The problem of finding the area bounded by the graph of a function under certain conditions.
These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus.
There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical tool for science and engineering
Definite integral: If a function f isdifferentiableinanintervalI, i.e., its derivative f ′ exists at each point of I, then a natural question arises that given f ′ at each point of I, can we determine the function? The functions that could possibly have given
a function as a derivative are called antiderivatives (or primitive) of the function. Further, the formula that gives all these antiderivatives is called the indefinite integral of the function and such process of finding antiderivatives is called integration.
Integration using substitution: this method uses the concept of change of variables in which we substitute some other function for the existing function to make the integration easy (by putting the other function as a substitution of the independent variable), given integrations is transformed to some easy integrations using and then integrated.
For example let then we substitute x = g(t)
Differentiating x = g(t) both sides we get dx = g′(t) dt
Integration by parts: this method is quite useful when we integrate product of functions If u and v are any two differentiable functions of a single variable x (say). Then, by the product rule of differentiation, we have
d (uv) = u dv + v du
Integrating both sides, we get
or
If we take f as the first function and g as the second function, then this formula may be stated as follows:
“The integral of the product of two functions = (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function)× (integral of the second function)]”
Integral of some particular functions: integral of the type
We have
Thus,
Definite integrals: The definite integral has a unique value. A definite integral is denoted by , where a is called the lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an antiderivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) – F(a)
Fundamental theorem of calculus:
First fundamental theorem of calculus: Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′(x) = f (x), for all x ∈ [a, b].
The second fundamental theorem of calculus: Let f be a continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then
.
Evaluation of definite integral by substitution: To evaluate , by substitution, the steps could be as follows:
Integral calculus is a very important topic to be understood. Integral calculus is one part of whole calculus other is differential calculus and it is used as a tool in physics, so you must be through with this chapter.
To understand this topic you have to go in the order of the name of the topic of this chapter means first you must study about what is integral and its type then notation of integration and then basic integration.
Start with understanding basic concepts like simple definitions (indefinite integral, the difference between indefinite and definite integral), understand each topic independently before moving to another by going through every concept.
Once you’re clear with basic concepts move to complex concepts, like properties of integration, different ways of integration.
After studying these concepts go through solved examples and then go to mcq and practice the problem to make sure you understood the topic.
Solve the questions of the books which you are following and then go to previous year papers.
While going through concept make sure you understand the derivation of formulas and try to derive them by your own, as many times you will not need the exact formula but some steps of derivation will be very helpful to solve the problem if you understand the derivation it will boost your speed in problemsolving.
At the end of chapter try to make your own short notes for quick revision, make a list of formula to revise quickly before exams or anytime when you required to revise the chapter, it will save lots of time for you.
First, finish all the concept, example and questions given in NCERT Maths Book. You must be thorough with the theory of NCERT. Then you can refer to the book Integral Calculus Arihant by Amit m. Agarwal or RD Sharma but make sure you follow any one of these not all. Integral calculus is explained very well in these books and there are an ample amount of questions with crystal clear concepts. Choice of reference book depends on person to person, find the book that best suits you the best, depending on how well you are clear with the concepts and the difficulty of the questions you require.
Chapters 
Chapters Name 
Chapter 1 

Chapter 2 

Chapter 3 

Chapter 4 

Chapter 5 

Chapter 6 

Chapter 7 

Chapter 9 

Chapter 10 

Chapter 11 

Chapter 12 

Chapter 13 

Chapter 14 

Chapter 15 

Chapter 16 
The integral
is equal to
(where C is a constant of integration)
The integral equals :
The area (in sq. units) of the smaller portion enclosed between the curves, x^{2}+y^{2}=4 and y^{2}=3x, is :