Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Quantitative Aptitude for CA CPT

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CA Foundation : Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

The document Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev is a part of the CA Foundation Course Quantitative Aptitude for CA CPT.
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9 The Fundamental Theorem in Terms of Differentials

Fundamental Theorem of Calculus: If F (x) is one antiderivative of the function f (x), i.e., F'(x) = f (x), then

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Thus, the integral of the diffierential of a function F is equal to the function itself plus an arbitrary constant. This is simply saying that diffierential and integral are inverse math operations of each other. If we rst dierentiate a function F(x) and then integrate the derivative F'(x) = f (x), we obtain F (x) itself plus an arbitrary constant. The opposite also is true. If we fi rst integrate a function f (x) and then dierentiate the resulting integral F (x) + C , we obtain F'(x) = f (x) itself.

Example:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example:Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example:Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

10 Integration by Subsitution

Substitution is a necessity when integrating a composite function since we cannot write down the antiderivative of a composite function in a straightforward manner.
Many students nd it difficult to gure out the substitution since for dierent functions the subsitutions are also different. However, there is a general rule in substitution, namely, to change the composite function into a simple, elementary function.

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: Note that sin(√x) is not an elementary sine function but a composite function. The rst goal in solving this integral is to change sin(√x) into an elementary sine function through substitution. Once you realize this, u = √x is an obvious subsitution. Thus, du = u'dx = 1/2√x dx,  or dx = 2√x du = 2udu. Substitute into the integral, we obtain

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Once you become more experienced with subsitutions and diffierentials, you do not need to do the actual substitution but only symbolically. Note that x = (√x)2,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Thus, as soon as you realize that √x is the substitution, your goal is to change the diffierential in the integral dx into the diffierential of px which is d√x.

If you feel that you cannot do it without the actual substitution, that is ne. You can always do the actual substitution. I here simply want to teach you a way that actual subsitution is not a necessity!

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: Note that cos(x5 ) is a composite function that becomes a simple cosine function only if the subsitution u = x5 is made. Since du = u'dx = 5x4 dx, xdx = 1/5du. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Or alternatively,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: Note that  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  is a composite function. We realize that u = x2 + 1 is a substitution. du = u'dx = 2xdx implies xdx = 1/2du. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Or alternatively,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

In many cases, substitution is required even no obvious composite function is involved.

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: The integrand ln x/x is not a composite function. Nevertheless, its antiderivative is not obvious to calculate. We need to gure out that (1/x)dx = d ln x, thus by introducing the substitution u = ln x, we obtained a diffierential of the function ln x which also appears in the integrand. Therefore,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

It is more natural to consider this substitution is an attempt to change the diffierential dx into something that is identical to a function that appears in the integrand, namely d ln x. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: tan x is not a composite function. Nevertheless, it is not obvious to gure out tan x is the derivative of what function. However, if we write tan x = sin x/cos x, we can regard 1/cos x as a composite function. We see that u = cosx is a candidate for substitution and du = u'dx = sin(x)dx. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev    Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Or alternatively,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Some substitutions are standard in solving speci c types of integrals.

Example: Integrands of the type  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

In this case both x = asinu and x = acosu will be good. x = a tanh u also works (1 tanh2 u = sech2 u). Let's pick x = asinu in this example. If you ask how can we nd out that x = asinu is the substitution, the answer is a2 - x2 = a2(1 - sin2u) = a2 cos2u. This will help us eliminate the half power in the integrand. Note that with this substitution, u = sin-1(x/a), sinu = x/a, and cosu = √1 - x2/a2

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev , we can make a simple substitution x = au, thus

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Similarly,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

x = asinh(u) is a good substitution since a2 + x2 = a2 + a2sinh2 (u) = a2 (1 + sinh(u)) = a2 cosh2(u), where the hyperbolic identity 1 + sinh2 (u) = cosh2(u) was used. (x = a tan u is also good since 1 + tanu = sec2u!). Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev    Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

where the following hyperbolic identities were used: sinh(2u) = 2sinh(u)cosh(u), sinh-1(x/a) =  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example: Integrands of the type  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

x = acosh(u) is prefered since x2 a= a2scos2 (u) a2 = a2[cosh2 (u) 1] = asinh(u), where the hyperbolic identity cosh2 (u) 1 = sinh2 (u) was used. (x = a sec u is also good since secu 1 = tanu!). Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Similarly

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev    Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

where cosh u = x/a, sinhu  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

1. Substitution aimed at eliminating a composite function

Example:

1. Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

2.   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

2. Substitution to achieve a function in diffierential that appears in the integrand

Example:

1.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

2.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev 

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

3. Special Trigonometric Substitutions 

Example:

1.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

11 Integration by Parts

Integration by Parts is the integral version of the Product Rule in dierentiation. The Product Rule in terms of diffierentials reads,

d(uv) = vdu + udv

Integrating both sides, we obtain

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

now that  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev , the above equation can be expressed in the following form,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Generally speaking, we need to use Integration by Parts to solve many integrals that involve the product between two functions. In many cases, Integration by Parts is most efficient in solving integrals of the product between a polynomial and an exponential, a logarithmic, or a trigonometric function. It also applies to the product between exponential and trigonometric functions.

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solutions: In order to eliminate the power function x, we note that (x)' = 1. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  = xex - ex + C

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: In order to eliminate the power function x2 , we note that (x2 )" = 2. Thus, we need to use Integration by Patrs twice.

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev 

= x2sinx + 2xcosx - 2sinx + C

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solution: In order to eliminate the power function x2 , we note that (x2 )"= 2. Thus, we need to use Integration by Patrs twice. However, the number (-2) can prove extremely annoying and easily cause errors. Here is how we use substitution to avoid this problem.

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev    Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev 

  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

When integrating the product between a polynomial and a logarithmic function, the main goal is to eliminate the logarithmic function by dierentiating it. This is because (ln x)' = 1/x.

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solutions:  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solutions:   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

More Exercises on Integration by Parts: 

Example:

1.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

2.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev 

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

12 Integration by Partial Fractions

Rational functions are defined as the quotient between two polynomials:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

where Pn(x) and Qm (x) are polynomials of degree n and m respectively. The method of partial fractions is an algebraic technique that decomposes R(x) into a sum of terms:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

where p(x) is a polynomial and Fi(x); (i = 1; 2; ..... ; k) are fractions that can be integrated easily.
The method of partial fractions is an area many students nd very difficult to learn. It is related to algebraic techniques that many students have not been trained to use. The most typical claim is that there is no fixed formula to use. It is not our goal in this course to cover this topics in great details (Read Edwards/Penney for more details). Here, we only study two simple cases.

Case I: Qm(x) is a power function, i.e., Qm(x) = (x - a)(Qm (x) = xm if a = 0!).

Example:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

This integral involves the simplest partial fractions:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Some may feel that it is easier to write the fractions in the following form: D-1(A + B + C ) = D-1A + D-1B + D-1C . Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Solutions: 

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Case II: Pn(x) = A is a constant and Qm (x) can be factorized into the form Q2 (x) = (x - a)(x - b), Q3 (x) = (x- a)(x - b)(x - c), or Qm (x) = (x - a1 )(x - a2) ..... (x - am )

Example:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Since, Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev  implies that A(x + 1) + B (x - 3) = 1. Setting x = 3 in this equation, we obtain A = 1/4. Setting x = -1, we obtain B = -1/4. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Example:

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Since  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev implies that A(x - 2)(x 3) + B (x -1)(x - 3) + C (x - 1)(x - 2) = 1. Setting x = 1 in this equation, we obtain A = 1/2. Setting x = 2, we obtain B = -1. Setting x = 3, we get C = 1/2. Thus,

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev   Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

More Exercises on Integration by Partial Fractions: 

Examples:

1.  Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

2. Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

3. Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

Differential and Integral Calculus (Part - 3) CA Foundation Notes | EduRev

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