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Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals : CBSE is a renowned educational Board, which comes under the Union Government of India. This eminent board was formed in 1952 and associated with the Board of High School and Intermediate Education, Rajputana. Ajmer, Gwalior, Merwara and Central India were included in the administrative territory of this board along with the other places including Bhopal, Ajmer and Vindhya Pradesh. From 1952 onwards, it has been providing a standard education and robust learning environment to all. The Central Board of Secondary Education or CBSE is a prestigious board of education and it provides affiliation to public and private schools. Apart from this, all Jawahar Navodaya Vidyalayas and kendriya vidyalayas are affiliated to this board.

Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals : Here we provides you Karnataka Class 12 Commerce Maths Integrals Complete Notes in PDF Format. Karnataka Class 12 Commerce Maths Integrals topic are  Integration as inverse process of differentiation: List of all the results immediately follows from knowledge of differentiation. Geometrical Interpretation of indefinite integral, mentioning elementary properties and problems. Methods of Integration: Integration by substitution, examples. Integration using trigonometric identities, examples, Integration by partial fractions: problems related to reducible factors in denominators only. Integrals of some particular functions : Evaluation of integrals of ∫ , ∫ √  ∫ √ and problems . Problems on Integrals of functions like ∫ , ∫ √ Integration by parts : Problems , Integrals of type ∫ [ ( ) ( )] and related simple problems. Evaluation of Integrals of some more types like √ , √ and problems Definite integrals: Definition, Definite Integral as a limit of a sum to evaluate integrals of the form ∫ ( ) only. Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals These all topics covers.

Download here Karnataka Class 12 Commerce Maths Complete Notes In PDF Format 

Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals : In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus, with its inverse, differentiation, being the other. Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral

{\displaystyle \int _{a}^{b}\!f(x)\,dx}\int _{a}^{b}\!f(x)\,dx

is defined informally as the signed area of the region in the xy-plane that is bounded by the graph of f, the x-axis and the vertical lines x = a and x = b. The area above the x-axis adds to the total and that below the x-axis subtracts from the total.

Roughly speaking, the operation of integration is the reverse of differentiation. For this reason, the term integral may also refer to the related notion of the antiderivative, a function F whose derivative is the given function f. In this case, it is called an indefinite integral and is written:

{\displaystyle F(x)=\int f(x)\,dx.}F(x)=\int f(x)\,dx.

The integrals discussed in this article are those termed definite integrals. It is the fundamental theorem of calculus that connects differentiation with the definite integral: if f is a continuous real-valued function defined on a closed interval [a, b], then, once an antiderivative F of f is known, the definite integral of f over that interval is given by

{\displaystyle \int _{a}^{b}\!f(x)\,dx=F(b)-F(a).}\int _{a}^{b}\!f(x)\,dx=F(b)-F(a).

The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century, who thought of the integral as an infinite sum of rectangles of infinitesimal width. A rigorous mathematical definition of the integral was given by Bernhard Riemann. It is based on a limiting procedure that approximates the area of a curvilinear region by breaking the region into thin vertical slabs. Beginning in the nineteenth century, more sophisticated notions of integrals began to appear, where the type of the function as well as the domain over which the integration is performed has been generalised. A line integral is defined for functions of two or three variables, and the interval of integration [a, b] is replaced by a certain curve connecting two points on the plane or in the space. In a surface integral, the curve is replaced by a piece of a surface in the three-dimensional space.

Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals : The definite integral is defined to be exactly the limit and summation that we looked at in the last section to find the net area between a function and the x-axis.  Also note that the notation for the definite integral is very similar to the notation for an indefinite integral.  The reason for this will be apparent eventually.

There is also a little bit of terminology that we should get out of the way here.  The number “a” that is at the bottom of the integral sign is called the lower limit of the integral and the number “b” at the top of the integral sign is called the upper limit of the integral.  Also, despite the fact that a and b were given as an interval the lower limit does not necessarily need to be smaller than the upper limit.  Collectively we’ll often call a and b the interval of integration.

Let’s work a quick example.  This example will use many of the properties and facts from the brief review of summation notation in the Extras chapter.

Example 1  Using the definition of the definite integral compute the following.


First, we can’t actually use the definition unless we determine which points in each interval that well use for .  In order to make our life easier we’ll use the right endpoints of each interval.

From the previous section we know that for a general n the width of each subinterval is,

The subintervals are then,

As we can see the right endpoint of the ith subinterval is

The summation in the definition of the definite integral is then,

Now, we are going to have to take a limit of this.  That means that we are going to need to “evaluate” this summation.  In other words, we are going to have to use the formulas given in the summation notation review to eliminate the actual summation and get a formula for this for a general n.

To do this we will need to recognize that n is a constant as far as the summation notation is concerned.  As we cycle through the integers from 1 to n in the summation only i changes and so anything that isn’t an i will be a constant and can be factored out of the summation.  In particular any n that is in the summation can be factored out if we need to.

Here is the summation “evaluation”.

We can now compute the definite integral.

We’ve seen several methods for dealing with the limit in this problem so I’ll leave it to you to verify the results.

Wow, that was a lot of work for a fairly simple function.  There is a much simpler way of evaluating these and we will get to it eventually.  The main purpose to this section is to get the main properties and facts about the definite integral out of the way.  We’ll discuss how we compute these in practice starting with the next section.

Karnataka Class 12 Commerce Maths Integrals Complete Notes

Karnataka Class 12 Commerce Maths Integrals :  See the Proof of Various Integral Properties section of the Extras chapter for the proof of properties 1  4.  Property 5 is not easy to prove and so is not shown there.  Property 6 is not really a property in the full sense of the word.  It is only here to acknowledge that as long as the function and limits are the same it doesn’t matter what letter we use for the variable.  The answer will be the same.

Let’s do a couple of examples dealing with these properties.

Example 2  Use the results from the first example to evaluate each of the following.





All of the solutions to these problems will rely on the fact we proved in the first example.  Namely that,


In this case the only difference between the two is that the limits have interchanged.  So, using the first property gives,


For this part notice that we can factor a 10 out of both terms and then out of the integral using the third property.


In this case the only difference is the letter used and so this is just going to use property 6.

Example 5  Given that , , and  determine the value of .


This example is mostly an example of property 5 although there are a couple of uses of property 1 in the solution as well.

We need to figure out how to correctly break up the integral using property 5 to allow us to use the given pieces of information.  First we’ll note that there is an integral that has a “-5” in one of the limits.  It’s not the lower limit, but we can use property 1 to correct that eventually.  The other limit is 100 so this is the number c that we’ll use in property 5.

We’ll be able to get the value of the first integral, but the second still isn’t in the list of know integrals.  However, we do have second limit that has a limit of 100 in it.  The other limit for this second integral is -10 and this will be c in this application of property 5.

At this point all that we need to do is use the property 1 on the first and third integral to get the limits to match up with the known integrals.  After that we can plug in for the known integrals.

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