Class 12 Chapter 8 (Application of Integrals) class notes

  Application of Integrals

Introduction

Calculus is widely seen as functions, differentiation, and integration. It has a lot of applications in the fields of Engineering, Science, and Mathematics. We have already studied to find the area bounded by the curve y = f(x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum.

                         

 

In this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses. We will learn how to use integrals to find out areas under different conditions.

Area under Simple Curves

Let we want to find the area bounded be the curve y = f(x), x-axis and the ordinates x = a and x = b.

From the figure, we assume that area under the curve is composed of large number of very thin vertical strips. Let us take an arbitrary strip of height y and width dx, then dA (area of the elementary strip) = y dx, where, y = f(x). This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b.

Total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f(x) is calculated by adding up the elementary areas of thin strips across the region PQRSP.

The area A under the curve f(x) bounded by x = a and x = b is given by

A = _aꭍ^b dA = _aꭍ^b y dx = _aꭍ^b f(x) dx

The area A of the region bounded by the curve x = g (y), y-axis 

and the lines y = c, y = d as shown in the figure, is given by

A = _cꭍ^d x dy = _cꭍ^d g(y) dy

Example: Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.

Solution:

The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

So, area of ABCD = ʃ₁⁴ y dx

                               = ʃ₁⁴ √x dx

                               = [x^3/2/(3/2)]₁⁴ 

                               = (2/3)[4^3/2 - 1^3/2]

                               = (2/3)[8 – 1]

                               = (2/3) * 7

                               = 14/3 units

Example: Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Solution:

The area of the region bounded by the curve, x² = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

So, area of ABCD = ʃ₂⁴ x dy

                               = ʃ₂⁴ 2√y dy

                               = 2[y^3/2/(3/2)]₂⁴ 

                               = (4/3)[4^3/2 - 2^3/2]

                               = (4/3)[8 – 2√2]

     p;                         = (32 – 8√2)/3 units

The area of the region bounded by a curve and a line

In this section, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse when their standard forms are given.

Example: Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle  x² + y² = 4

Solution:

The area of the region bounded by the circle x² + y² = 4, x = √3y, and the x-axis is the  area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1).

Area OAB = Area ∆OCA + Area ACB

Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2

Area ACB = ʃ_√3² y dx

                  = ʃ_√3² √(4 – x²) dx

                  = [x√(4 – x²)/2 + (4/2) * sin^-1 x/2]_√3² 

                  = [2 * π/2 - (√3/2) * √(4 - 3) - 2sin^-1 √3/2]

                  = [π - √3/2 – 2(π/3)]

                  = [π/3 - √3/2]

Therefore, area enclosed by x-axis, x = √3y and the circle x² + y² = 4 in the first quadrant

= √3/2 + [π/3 - √3/2] = π/3 units

Example: Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = a/√2

Solution:

The area of the smaller part of the circle, x² + y² = a² cut off by the line x = a/√2 is the area ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis.

                                                        

So, Area ABCD = 2 * Area ABC

Now, Area of ABC = ʃ_a/√2^a y dx

                                 = ʃ_a/√2^a √(a² – x²) dx

                                 = [x√( a² – x²)/2 + (a²/2) * sin^-1 x/a]_a/√2^a 

                                 = [a²/2 * π/2 - (a/2√2) * √( a² - a²/2) - a²/2 * sin^-1 1/√2]

                                 = a²/2 * π/2 - (a/2√2) * (a/√2) - a²/2 * (π/4)

                                  = a²π/4 - a²/4 - a²π/8

                                  = (a²/4)(π - 1 - π/2)

                                  = (a²/4)(π/2 - 1)          

So, area of ABCD = 2[(a²/4)(π/2 - 1)] = (a²/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x² + y² = a² cut off by the line x = a/√2 is

(a²/2)(π/2 - 1) units.

 

Area between Two Curves

The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. Let we are given two curves represented by y = f(x), y = g(x), where f(x) ≥ g(x) in [a, b] as shown in figure.

                                                

Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.

Elementary strip has height f(x) – g(x) and width dx so that the elementary area

dA = [f(x) – g(x)] dx

Total area A can be calculated as A = _aꭍ^b [f(x) – g(x)]

Alternatively,

A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]

– [area bounded by y = g (x), x-axis and the lines x = a, x = b]

A = _aꭍ^b f(x) = _aꭍ^b g(x) = _aꭍ^b [f(x) – g(x)], where f(x) ≥ g(x) in [a, b]

If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the figure, then the area of the regions bounded by curves can be calculated as

Total Area = Area of the region ACBDA + Area of the region BPRQB.

                   = _aꭍ^c [f(x) – g(x)] + _cꭍ^b [g(x) – f(x)]  

                            

Example:

Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

Solution:

The required area is represented by the shaded area OBCDO.

                                                  

Solving the given equation of circle, 4x² + 4y² = 9, and parabola, x² = 4y, we get the point of intersection as B(√2, 1/2) and D(-√2, 1/2).

It can be observed that the required area is symmetrical about y-axis.

So, Area OBCDO = 2 * Area OBCO

Now, draw BM perpendicular to OA.

Therefore, the coordinates of M is (√2, 0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

                                         = ʃ₀^√2  {√(9 – 4x²)/4} dx - ʃ₀^√2 x²/4 dx

                                         = (1/2)ʃ₀^√2 √(9 – 4x²) dx – (1/4)ʃ₀^√2 x² dx

                                        = (1/4)[x√(9 – 4x²) + (9/2) * sin^-1 2x/3]₀^√2  - (1/4)[x³/3]₀^√2            

                                        = (1/4)[√2 * √(9 – 8) + (9/2) * sin^-1 2√2/3] - (1/12)(√2)³

                                        = (√2/4) + (9/8) * sin^-1 2√2/3 - √2/6

                                        = (√2/12) + (9/8) * sin^-1 2√2/3

                                        = (1/2)[√2/6 + (9/4) * sin^-1 2√2/3]

Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin^-1 2√2/3]

                                                                   = [√2/6 + (9/4) * sin^-1 2√2/3] units

Example:

Find the area between the curves y = x and y = x²

Solution:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x² is A (1, 1).

Now, draw AC perpendicular to x-axis.

So, Area (OBAO) = Area (∆OCA) – Area (OCABO)

                              = ʃ₀¹ x dx - ʃ₀¹ x² dx

                              = [x²/2]₀¹ – [x³/3]₀¹

                              = 1/2 - 1/3

                              = 1/6 units

 

 

 

 

 

 

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Class 12 chapter 7 (Integrals) class notes

 

Introduction

Integral is one of the important concepts of calculus. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions as shown in the figure.

                                                                       

It is also used to find volumes, central points and many useful things.

The functions that could possibly have given function as a derivative are called anti derivatives or primitive of the function. Again, the formula that gives all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration.

The development of integral calculus is used to solve the problems of the following types:

(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. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability.

 

Integration as an Inverse Process of Differentiation

Integration is the inverse process of differentiation. Let f(x) be a function. Then the collection of all primitives is called the indefinite integral of f(x) and denoted by ʃ f(x) dx.

Thus, d[ф(x) + C]/dx = f(x) => ʃ f(x) dx = ф(x) + C

Where ф(x) is primitive of f(x) and C is an arbitrary constant known as constant of integration.

                                

The following is a list to find integrals of other functions.

                                    

                             

Some properties of indefinite integral

(i) The process of differentiation and integration are inverses of each other in the sense of the following results :

d[ʃ f(x) dx]/dx = f(x)

and  ʃ f’(x) dx = f(x) + C, where C is any arbitrary constant.

(ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent.

i.e. d[ʃ f(x) dx]/dx = d[ʃ g(x) dx]/dx

=> d[ʃ f(x) dx - ʃ g(x) dx]/dx = 0

=> ʃ f(x) dx - ʃ g(x) dx = C, where C is any real number

=> ʃ f(x) dx = ʃ g(x) dx + C

So, the families of curve [ʃ f(x) dx + C₁, C₁ є R] and [ʃ g(x) dx + C₂, C₂ є R] are identical.

Hence, ʃ f(x) dx and ʃ g(x) dx are equivalent.

(iii) ʃ [f(x) + g(x)] dx = ʃ f(x) dx + ʃ g(x) dx

(iv) For any real number k, ʃ [k * f(x)] dx = k * ʃ f(x) dx

 (v) ʃ [k₁ * f₁(x) + k₂ * f₂(x) + . . . . .+ k_n * f_n(x)] dx = k₁ * ʃ f₁(x) dx + k₂ * ʃ f₂(x) dx + . . . . . .+ k_n * ʃ f_n(x) dx

Problem: Find the integral of the following functions:

(a) ʃ (ax² + bx + c) dx  (b) ʃ (2x² + e^x) dx

Solution:

(a) ʃ (ax² + bx + c) dx = ʃ ax² dx + ʃ bx dx + ʃ c dx

                                     = aʃ x² dx + b ʃ x dx + c ʃ dx

                                     = ax³/3 + bx²/2 + cx + C

So, ʃ (ax² + bx + c) dx = ax³/3 + bx²/2 + cx + C

(b) ʃ (2x² + e^x) dx = ʃ 2x² dx + ʃ e^x dx

                              = 2ʃ x² dx + ʃ e^x dx

                              = 2x³/3 + e^x + C

So, ʃ (2x² + e^x) dx = 2x³/3 + e^x + C

Methods of Integration

There are certain methods for solving the integration:

1. Integration by Substitution
2. Integration using Partial Fractions
3. Integration by Parts
4. Integration by Substitution

The given integral ʃ f(x) dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t).

Let I = ʃ f(x) dx

Put x = g(t) so that dx/dt = g′(t).

We write dx = g′(t) dt

Thus I = ʃ f(x) dx = ʃ f{g(t)} * g’(t) dt

Problem: Integrate the following functions:

(a) 2x/(1 + x²)   (b) (log x)²/x

Solution:

(a) Let 1 + x² = t

=> 2x dx = dt

Now, ʃ 2x/(1 + x²) dx = ʃ dt/t

                                      = log|t| + C

                                      = log|1 + x²| + C

                                      = log(1 + x²) + C

(b) Let log x = t

=> dx/x = dt

Now, ʃ [(log x)²/x] dx = ʃ t² dt     

                                      = t³/3 + C

                                      = (log|x|)³/3 + C

There are some important trigonometries integral which are listed below:

(i) ʃ tan x dx = log|sec x| + C  

(ii) ʃ cot x dx = log|sin x| + C

(iii) ʃ sec x dx = log|sec x + tan x| + C

(iv) ʃ cosec x dx = log|cosec x – cot x| + C

Integrals of Some Particular Functions

                             

To find the integral ʃ dx/(ax² + bx + c)

We write ax² + bx + c = a[x² + bx/a + c/a] = a[(x + b/2a)² + (c/a – b²/4a²)]

Put x + b/2a = t so that dx = dt and write c/a – b²/4a² = ±k²

Now, ʃ dx/(ax² + bx + c) = (1/a) * ʃ dt/(t² ± k²)

If c/a – b²/4a² is +ve then we put + k² and if If c/a – b²/4a² is +ve then we put - k²

 

Problem: Find the integral

1/(9x² + 6x + 5)

Solution:

1/(9x² + 6x + 5) = 1/{(3x + 1)² + 2²}

Let 3x + 1 = t

=> 3 dx = dt

=> dx = dt/3

So, ʃ dx/(9x² + 6x + 5) = (1/3) * ʃ dt/(t² + 2²)

                                       = (1/3) * [(1/2) * tan^-1(t/2)] + C

                                       = (1/6) * tan^-1{(3x + 1)/2)} + C

To find the integral of type ʃ dx/√(ax² + bx + c):

To integrate ʃ dx/√(ax² + bx + c), we do the same process as to find the integral of type ʃ dx/(ax² + bx + c) and then apply the standard formula to obtain the solution.

Problem:  Find the integral

1/√(x² + 2x + 2)

Solution:

1/√(x² + 2x + 2) = 1/√{(x + 1)² + 1²}

Let x + 1 = t

=> dx = dt

So, ʃ dx/√(x² + 2x + 2) = ʃ dt/√(t² + 1²)

                                       = log[t + √(t² + 1)] + C

                                       = log[(x + 1) + √{(x + 1)² + 1)}] + C

                                       = log[(x + 1) + √(x² + 2x + 2)] + C

 

To find the integral of type ʃ [(px + q)/(ax² + bx + c)] dx where p, q, r and s are constant:

 To integrate ʃ [(px + q)/(ax² + bx + c)] dx, we have to find real numbers A and B such that

      px + q = A * d(ax² + bx + c)/dx + B

=> px + q = A(2ax + b) + B

To determine the value of A and B, we equate from both sides the coefficients of x and the constant terms. After finding A and B, the integral is reduced to one of the known forms.

 

Problem: Find the integral

 (4x + 1)/(2x² + x - 3)     

Solution:

Let 4x + 1 = A * d(2x² + x - 3)/dx + B

=> 4x + 1 = A(4x + 1) + B

=> 4x + 1 = 4Ax + A + B

Equating the coefficients of x and constant term on both sides, we get

4A = 4

=> A = 1

A + B = 1

=> 1 + B = 1

=> B = 0

Let 2x² + x – 3 = t

=> (4x + 1)dx = dt

Now, ʃ [(4x + 1)/(2x² + x - 3)]dx = ʃ dt/t

                                                       = log t + C

                                                       = log(2x² + x - 3) + C

 

To find the integral of type ʃ [(px + q)/√(ax² + bx + c)] dx where p, q, r and s are constant:

 To integrate ʃ [(px + q)/√(ax² + bx + c)] dx, we do the same process as to find the integral of type

 ʃ [(px + q)/(ax² + bx + c)] dx and transform the integral into known standard forms.

Problem: Find the integral

(5x + 3)/√(x² + 4x + 10)

Solution:

Let (5x + 3) = A * d(x² + 4x + 10)/dx + B

=> (5x + 3) = A(2x + 4) + B

Equating the coefficients of x and constant term on both sides, we get

2A = 5

=> A = 5/2

And 4A + B = 3

=> B = -7

So, (5x + 3) = 5(2x + 4)/2 – 7

Now, ʃ [(5x + 3)/√(x² + 4x + 10)]dx = ʃ [{5(2x + 4)/2 – 7}/√(x² + 4x + 10)]dx

                                                             = (5/2) * ʃ [(2x + 4)/√(x² + 4x + 10)]dx - 7 * ʃ dx/√(x² + 4x + 10)

Let I₁ = ʃ [(2x + 4)/√(x² + 4x + 10)]dx and I₂ = ʃ dx/√(x² + 4x + 10)

So, ʃ [(5x + 3)/√(x² + 4x + 10)]dx = 5I₁/2 – 7I₂    ……………..1

Now, consider I₁ = ʃ [(2x + 4)/√(x² + 4x + 10)]dx

Let x² + 4x + 10 = t

=> (2x + 4)dx = dt

So, I₁ = ʃ dt/√t

=> I₁ = 2√t

=> I₁ = 2√(x² + 4x + 10)

Again consider I₂ = ʃ dx/√(x² + 4x + 10)

=> I₂ = ʃ dx/√(x² + 4x + 4 + 6)

=> I₂ = ʃ dx/√{(x + 2)² + 6}

=> I₂ = ʃ dx/√{(x + 2)² + (√6)²}

=> I₂ = log{(x + 2) + (x² + 4x + 10)}

From equation 1, we get

      ʃ [(5x + 3)/√(x² + 4x + 10)]dx = 5[2√(x² + 4x + 10)]/2 – 7 * log{(x + 2) + (x² + 4x + 10)}

=> ʃ [(5x + 3)/√(x² + 4x + 10)]dx = 5 * √(x² + 4x + 10) – 7 * log{(x + 2) + (x² + 4x + 10)}

Integration by Partial Fractions

We know that a rational function is the ratio of two polynomials in the form P(x)/Q(x), where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If degree of P(x) is less than degree of Q(x), then it is called proper rational function otherwise it is improper. We can convert an improper rational function into proper rational function by division method.

 So, if P(x)/Q(x) is improper, then P(x)/Q(x) = T(x) + P₁(x)/Q(x)

Now to calculate ʃ [P(x)/Q(x)] dx where P(x)/Q(x) is a proper rational function, we first convert it into partial faction suing the method as shown in the given figure and then integrate using standard formula.

 Problem: Find the integral

x/{(x - 1)(x - 2)(x - 3)}

Solution:

Let x/{(x - 1)(x - 2)(x - 3)} = A/(x - 1) + B/(x - 2) + C/(x - 3)

=> x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

Equating the coefficients of x², x and constant term, we get

A + B + C = 0

-5A – 4B – 3C = 1

6A + 3B + 2C = 0

Solving these equations, we get

A = 1/2, B = −2, and C = 3/2

So, x/{(x - 1)(x - 2)(x - 3)} = 1/2(x - 1) – 2/(x - 2) + 3/2(x - 3)

Now, ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = ʃ dx/2(x - 1) – ʃ dx/2(x - 2) + ʃ 3dx/2(x - 3)

=> ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * ʃ dx/(x - 1) – 2 ʃ dx/(x - 2) + (3/2) * ʃ dx/(x - 3)

=> ʃ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * log|x – 1| – 2 log|x – 2| + (3/2) * log|x – 3| + C

Problem: Find the integral

x/{(x² + 1)(x - 1)}

Solution:

Let x/{(x² + 1)(x - 1)} = (Ax + B)/(x² + 1) + C/(x - 1)

=> x = (Ax + B)(x - 1) + C(x² + 1)

=> x = Ax² – Ax + Bx – B + Cx² + C

Equating the coefficients of x², x and constant terms, we get

A + C = 0

-A + B = 1

-B + C = 0

On solving these equations, we get

A = -1/2, B = 1/2 and C = 1/2

From equation 1, we get

x/{(x² + 1)(x - 1)} = (-x/2 + 1/2)/(x² + 1) + 1/2(x - 1)

x/{(x² + 1)(x - 1)} = (-x/2)/(x² + 1) + (1/2)/(x² + 1) + 1/2(x - 1)

Now, ʃ [x/{(x² + 1)(x - 1)} dx = (-1/2)ʃ [x/(x² + 1)] + (1/2)ʃ [1/(x² + 1)] + (1/2)ʃ dx/(x - 1)

                                                = (-1/4)ʃ [2x/(x² + 1)] dx + (1/2) * tan^-1 x + (1/2) * log|x - 1| + C

Let x² + 1 = t

=> 2x dx = dt

So, ʃ [2x/(x² + 1)] = ʃ dt/t

                               = log|t|

                               = log|x² + 1|

Now, ʃ [x/{(x² + 1)(x - 1)}dx = (-1/4) log|x² + 1| + (1/2)tan^-1 x + (1/2)log|x - 1| + C

                                                = (1/2)log|x - 1|- (1/4) log|x² + 1| + (1/2)tan^-1 x + C

 

Integration by Parts

Integration by parts means integration of product of two numbers. To find the integral of product of two functions, we use the following formula:

“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)]”

Let f(x) and g(x) are two functions. Assume f(x) is the first function and g(x) is the second function.

Now, ʃ [f(x) * g(x)] dx = f(x) * ʃ g(x) dx – ʃ {f’(x) * ʃ g(x) dx} dx

 

Problem: Find the integral

x * sin x

Solution:

Let I = ʃ x * sin x dx

Taking x as first function and sin x as second function and integrating by parts, we get

I = x * ʃ sin x dx - ʃ [dx/dx * ʃ sin x dx]dx

I = x * (-cos x) - ʃ 1 * (-cos x)]dx

I = -x * cos x + ʃ cos x dx

I = -x * cos x + sin x + C

Integral of the type ʃ e^x[f(x) + f’(x)] dx

 Let I = ʃ e^x[f(x) + f’(x)] dx

          = ʃ e^x * f(x) dx + ʃ e^x * f’(x) dx

 => I = I₁ + ʃ e^x * f’(x) dx, where I₁ = ʃ e^x * f(x) dx     ………….1

Taking f(x) and first function and e^x as second function in I₁ and integrate, we get

I₁ = f(x) * e^x - ʃ f’(x) e^x + C

 From equation 1, we get

=> I = f(x) * e^x - ʃ f’(x) e^x + ʃ e^x * f’(x) dx + C

=> I = e^x * f(x) + C

Problem: Solve the integral

e^x(sin x + cos x)

Solution:

Let I = ʃ e^x(sin x + cos x) dx

Again let f(x) = sin x

=> f’(x) = cos x

So, I = ʃ e^x{f(x) + f’(x)} dx

It is known that ʃ e^x{f(x) + f’(x)} dx = e^x f(x) + C

So, I = e^x sin x + C

Integrals of some more types

There are some special types of integrals that will be solve using integration by parts:

(i) ʃ √(x² – a²) dx      (ii) ʃ √(x² + a²) dx   (iii) ʃ √(a² – x²) dx

(i) Let I = ʃ √(x² – a²) dx

=> I = ʃ 1 * √(x² – a²) dx

Taking 1 as the second function and integrating by parts, we get

I = x * √(x² – a²) - ʃ [(x/2) * 2x/√(x² – a²)] dx

=> I = x * √(x² – a²) - ʃ [x²/√(x² – a²)] dx

=> I = x * √(x² – a²) - ʃ [(x² – a² + a²)/√(x² – a²)] dx

=> I = x * √(x² – a²) - ʃ [(x² – a² )/√(x² – a²)] dx - ʃ [a²/√(x² – a²)] dx

=> I = x * √(x² – a²) - ʃ √(x² – a²) dx - a² ʃ dx/√(x² – a²)]

=> I = x * √(x² – a²) - I - a² log|x + √(x² – a²)| + C

=> 2I = x * √(x² – a²) - a² log|x + √(x² – a²)| + C

=> I = (x/2) * √(x² – a²) – (a² /2) * log|x + √(x² – a²)| + C

=> ʃ √(x² – a²) dx = (x/2) * √(x² – a²) – (a²/2) * log|x + √(x² – a²)| + C

Similarly, we can prove that

(ii) ʃ √(x² + a²) dx = (x/2) * √(x² + a²) + (a²/2) * log|x + √(x² + a²)| + C

(iii) ʃ √(a² – x²) dx = (x/2) * √(a² – x²) + (a²/2) * sin^-1 (x/a) + C

Problem: Find the integral

(i) √(1 – 4x²)      (ii) √(x² + 4x + 6)

Solution:

(i) Let I = √(1 – 4x²) = √{1² – (2x)²}

Again let 2x = t

=> 2 dx = dt

So, I = (1/2) √(1² – t²)

We know that ʃ √(a² – x²) dx = (x/2) * √(a² – x²) + (a²/2) * sin^-1(x/a) + C

So, I = (1/2)[(t/2) * √(1 – t²) + (1/2) * sin^-1 t] + C

=> I = (t/4) * √(1 – t²) + (1/4) * sin^-1 t + C

=> I = (2x/4) * √(1 – 4x²) + (1/4) * sin^-1 2x + C

=> I = (x/2) * √(1 – 4x²) + (1/4) * sin^-1 2x + C

(ii) Let I = ʃ √(x² + 4x + 6) dx

=> I = ʃ √(x² + 4x + 4 + 2) dx

=> I = ʃ √{(x² + 4x + 4) + 2} dx

=> I = ʃ √{(x + 2)² + (√2)²} dx

We know that ʃ √(x² + a²) dx = (x/2) * √(x² + a²) + (a²/2) * log|x + √(x² + a²)| + C

So, I = {(x + 2)/2} * √(x² + 4x + 6) + (2/2) * log|(x + 2) + √(x² + 4x + 6)| + C

=> I = {(x + 2)/2} * √(x² + 4x + 6) + log|(x + 2) + √(x² + 4x + 6)| + C

 

 

Definite Integral

A definite integral is denoted by _aʃ^b f(x) dx, 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 anti derivative 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).

 

Definite integral as the limit of a sum:

The definite integral _aʃ^b f(x) dx is the area bounded by the curve y = f (x), the ordinates x = a, x = b and the x-axis. To find this area Let us consider the region PQRSP between this curve and the ordinates x = a and x = b as shown in the given figure.

Now, divide the interval [a, b] into n equal sub intervals denoted by [x₀, x₁], [x₁, x₂], ……..,[x_r-1, x_r], ………..

 [x_n-1, x_n] where x₀ = a, x₁ = a + h, x₂ = a + 2h, ……….,x_r = a + rh, x_n = b = a + nh or h = (b - a)/n

 Again as n -> ∞, h -> 0

 The region PRSQP under consideration is the sum of n sub regions, where each sub region is defined on sub intervals [x_{r – 1}, x_r], r = 1, 2, 3, …, n.

Now, to find the area of the region PQRSQ is calculated as

Lim_x->∞ S_n = area of the region PQRSQ = _aʃ^b f(x) dx

   = lim_h->0 h[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

   = (b - a) * lim_h->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

 So, _aʃ^b f(x) dx = (b - a) * lim_h->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]

 Where h = (b - a)/n as n -> ∞

 This is known as the definition of definite integral as the limit of sum.

Problem: Evaluate the definite integral as limit of sums

ʃ₀⁵ (x + 1) dx

Solution:

We know that

ʃ_a^b f(x) dx = (b - a)lim_n->∞ (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n

Here a = 0, b = 5 and f(x) = x + 1

So, h = (5 - 0)/n = 5/n

Now, ʃ₀⁵ (x + 1) dx = (5 - 0)* lim_n->∞ (1/n)[f(0) + f(5/h) + …………..+ f{5(n - 1)/n}]

                                 = 5 * lim_n->∞ (1/n)[1 + (5/n + 1) + …………..+ {1 + 5(n - 1)/n}]

                                 = 5 * lim_n->∞ (1/n)[(1 + 1 + 1 + …….n times) + {5/n + 2 * 5/n + …………..+ 5(n - 1)/n}]

                                 = 5 * lim_n->∞ (1/n)[(1 + 1 + …….n times) + (5/n){1 + 2 + 3 +…………..+ (n - 1)}]

                                 = 5 * lim_n->∞ (1/n)[n + (5/n) * n(n - 1)/2]

                                 = 5 * lim_n->∞ (1/n)[n + 5(n - 1)/2]

                                 = 5 * lim_n->∞ (n/n)[1 + 5(1 – 1/n)/2]

                                 = 5 * [1 + 5(1 – 1/∞)/2]

                                 = 5 * [1 + 5(1 – 0)/2]

                                 = 5 * [1 + 5/2]

                                 = 5 * (7/2)

                                 = 35/2

Fundamental Theorem of Calculus

There are some fundamental theorems which are going to discuss here one by one.

Area function

Let x be the given point in [a, b] as shown in the figure.

 

Then _aʃ^b f(x) dx represents the area of the shaded region. The function A(x) is represented as Area function and is calculated as

A(x) = _aʃ^b f(x) dx

According to this definition, two fundamental theorems are stated as:

First fundamental theorem of integral calculus

Theorem 1: 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].

Second fundamental theorem of integral calculus

Theorem 2: Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then _aʃ^b f(x) dx = [F(x) _a]^b = F(b) – F(a), where F(x) is the indefinite integral of f(x).

Problem: Evaluate the following definite integrals

(a) ʃ_-1¹ (x + 1) dx                                                         (b) ʃ₂¹ dx/x

Solution:

(a) Let I = ʃ_-1¹ (x + 1) dx

Now, ʃ (x + 1) dx = x²/2 + x = F(x)

By second fundamental theorem of calculus, we get

I = F(1) - F(-1)

  = (1/2 + 1) – (1/2 - 1)

  = 1/2 + 1 – 1/2 + 1

  = 2

So, ʃ_-1¹ (x + 1) dx = 2

(b) Let I = ʃ₂¹ dx/x

Now, ʃ dx/x = log|x| = F(x)

By second fundamental theorem of calculus, we get

I = F(3) - F(2)

  = log|3| - log|2|

  = log 3 – log 2

  = log(3/2)

So, ʃ₂¹ dx/x = log(3/2)

 

Evaluation of Definite Integrals by Substitution

To evaluate _aʃ^b f(x) dx by substitution, the steps could be as follows:

1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.

2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.

3. Resubstitute for the new variable and write the answer in terms of the original variable.
4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the

values at the upper and lower limits.

Problem: Evaluate the following integrals using substitution method

(a) ʃ₀¹ [x/(x² + 1)] dx      (b) ʃ₀^π/2 [√(sin ф) * cos⁵ ф] dф

Solution:

(a) Let x² + 1 = t

=> 2x dx = dt

=> x dx = dt/2

When x = 0, t = 1 and when x = 1, t = 2

So, ʃ₀¹ [x/(x² + 1)] dx = (1/2)ʃ₁² dt/t

                                     = (1/2)[log t]₁²

                                     = (1/2)[log 2 – log 1]

                                     = (log 2)/2

 

(b) Let I = ʃ₀^π/2 [√(sin ф) * cos⁵ ф] dф

               = ʃ₀^π/2 [√(sin ф) * cos⁴ ф * cos ф] dф

Let sin ф = t

=> cos ф dф = dt

When ф = 0, t = 0 and when ф = π/2, t = 1

So, I = ʃ₀¹ [√t * (1 – t²)²] dt

        = ʃ₀¹ [√t * (1 + t⁴ - 2t²)] dt

        = ʃ₀¹ [t^1/2 + t^9/2 - 2t^5/2] dt

        = [t^3/2/(3/2) + t^11/2/(11/2) - 2t^7/2/(7/2)]₀¹

        = 1/(3/2) + 1/(11/2) - 2/(7/2)

        = 2/3 + 2/11 – 4/7

        = (154 + 42 - 132)/231

        = 64/231

 

 

Some Properties of Definite Integrals

There are some properties of definite integral which are very useful for calculating definite integral very

easily.

(i) _aʃ^b f(x) dx = _aʃ^b f(t) dt

(ii) _aʃ^b f(x) dx = -_bʃ^a f(t) dt

In particular, _aʃ^a f(x) dx = 0

(iii) _aʃ^b f(x) dx = _aʃ^c f(x) dx + _cʃ^b f(x) dx

(iv) _aʃ^b f(x) dx = _aʃ^b f(a + b - x) dx

(v) ₀ʃ^a f(x) dx = ₀ʃ^a f(a - x) dx

(vi) ₀ʃ^2a f(x) dx = ₀ʃ^a f(x) dx + ₀ʃ^a f(2a - x) dx

(vii) ₀ʃ^2a f(x) dx = 2 * ₀ʃ^a f(x) dx, if f(2a - x) = f(x)

                          = 0, if f(2a - x) = -f(x)

(vii) _-aʃ^a f(x) dx = 2 * ₀ʃ^a f(x) dx, if f is an even function i.e. f(-x) = f(x)

                          = 0, if f is an odd function i.e. f(-x) = -f(x)

Problem: Evaluate the following integrals

(a) ʃ₀^π/2 cos² x dx                                                                           (b) ʃ_-π/2^π/2 sin² x dx

Solution:

(a) Let I = ʃ₀^π/2 cos² x dx    …………..1

=> I = ʃ₀^π/2 cos² (π/2 - x) dx                              [Since ʃ₀^a f(x) dx = ʃ₀^a f(a - x) dx]

=> I = ʃ₀^π/2 sin² x dx      …………..2                           

Adding equation 1and 2, we get

      2I = I = ʃ₀^π/2 (sin² x + cos² x) dx

=> 2I = I = ʃ₀^π/2 1 dx

=> 2I = I = [x]₀^π/2

=> 2I = π/2

=> I = π/4

So, ʃ₀^π/2 cos² x dx = π/4

(b) Let I = ʃ_-π/2^π/2 sin² x dx

Since sin² (−x) = {sin(−x)}² = (−sin x)² = sin² x, therefore, sin² x is an even function.

It is known that if f(x) is an even function, then ʃ_-a^a f(x) dx = 2 * ʃ₀^a f(x) dx

So, I = 2 * ʃ₀^π/2 sin² x dx

=> I = 2 * ʃ₀^π/2 {(1 – cos 2x)/2} dx

=> I = ʃ₀^π/2 (1 – cos 2x) dx

=> I = [x – sin 2x /2]₀^π/2

=> I = [π/2 – sin (2 * π/2) /2]

=> I = [π/2 – (sin π)/2]

=> I = π/2

So, ʃ_-π/2^π/2 sin² x dx = π/2

Integrals

 

 

 

 

 

 

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