Basic Integration- II | Mathematics for Competitive Exams PDF Download

Integrals of the form

(i)

(ii)

(iii)

(iv)

(v)

(vi)

where a, b, c ∈ R and not at a time all zero.

Rule : We shall always in such cases divide above and below by cos2 x; then put tan x = t i.e., sec² x dx = dt then the questions shall reduce to the forms

Example 39 : Evaluate :

Let I =

=

=

Divide above and below by cos²x, then

=

Put √3 tan x = t

=

=

=

=

Integrals of the form

(i)

(ii)

(iii)

Rule : Divide above and below by cos² x, then 

Put a sec x + b tan x = t

Example 40 : Evaluate : 

Let I =

Divide above and below by cos² x, then

 

Put 2 sec x + 5 tan x = t

∴ (2 sec x tan x + 5 sec2 x) dx = dt

∴

=

=

(ii)

Rule : Divide above and below by sin² x, then 

Put a cosec x + b cot x = t

Example 41: Evaluate :

Let I =

Divide above and below by sin² x, then

Put cosec x + 2 cot x = t

∴ (- cosec x cot x - 2 cosec² x) dx = dt

or (cosec x cot x + 2 cosec² x) dx = - dt 

then

=

=

Integrals of the form

(i)

(ii) 

(i)

Rule : Since integral for sin x + cos x is sin x - cos x 

Put sinx - cos x = t  ∴(cos x + sin x) dx = dt 

and sin² x + cos2 x - 2 sin x cos x = t² or sin 2x = 1 - t²

then  it can be evaluated easily. 

Example 42 : Evaluate : 

Let I =

=

=

Put sin x - cos x = t ∴(cos x + sin x) dx = dt

and sin² x + cos² x - 2 sin x cos x = t²

⇒

then

=

=

(ii) 

Rule : Since integral of sin x - cos x is - cos x - sin x 

Put sin x + cos x = t (cos x - sin x) dx = dt 

and sin² x + cos² x + 2 sin x cos x = t² 

or sin 2x = t² - 1

then  it can be evaluated easily.

Example 43 : Evaluate : 

Let I =

Put sin x + cos x = t (cos x - sin x) dx = dt

⇒ (sin x - cos x) dx = - dt 

and sin² x + cos² x + 2 sin x cos x = t²

⇒

then

=

=

Again put t⁴ - 1 = z²

⇒ 4t³ dt = 2z dz or 2t³ dt = z dz

then

=

=

Integral of the form

Rule : 

(i) If R ( - sin x, cos x) = - R (sin x, cos x) then put cos x = t 

(ii) If R (sin x, - cos x) = - R (sin x, cos x) then put sin x = t 

(iii) If R ( - sin x, - cos x) = R (sin x, cos x) then put tan x = t

Example 44 : Evaluate :

Let I =

Let

⇒

therefore we put cos x = t ∴ sin x dx = - dt

∵

=

=

=

=

=

=

=

Integrals of the form

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Rule for (i), (ii), (iii), (iv), (v) :

Write cos x =

the numerator will become sec²(x/2) and the denominator will be a quadratic in tan(x/2). Putting tan(x/2) = t i.e., sec²(x/2) dx = 2 dt the question will reduces to the form

Example 45 : Evaluate :

Let I =  =

=

Put tan (x/2) = t   ∴ sec² (x/2) dx = 2 dt

then I =

=

Integrals of the form

(i)

(ii) 

(i) 

Rule : Put t =                     -----(1)

 or  -----(2)

Example 46 : Evaluate 

Put  -----(1)

∴  -----(2)

From (1), sin x = then cos x =

From (2),

Integrating both sides, we get

Let 1 - t² = z²⇒ - tdt = zdz

=

=

=

=

=      

(ii) 

Rule : Put t =     ------(1)

∴  or

Now substitute the value of sin x in L.H.S. from (1) in terms of t and then integrate both sides.

Example 47 : Evaluate : 

Let I =  

Put     ------(1)

∴  or  ------(2)

From (1),

∴

From (2),

Integrating both sides, we get

=

=

Integrals of the form 

(i)

(ii)

(iii)

(iv)

(i) 

Rule :

Let I =   =

=

=

In first integral on R.H.S. put tan x = t and in second integral put sin x = u 

Example 48 : Evaluate : 

 Let I =

=

=

=

=

In first integral on R.H.S. put tan x = t and in second integral put √sin x = u.

i.e., sec² x dx = dt and cos x dx = du/√2

=

=

(iii)  and

(iv) 

Rule : In this case change tan² x into sec² x - 1 and cot² x into cosec² x - 1 and then proceed as in (i) and (ii).

Integrals of the form

(i)

(ii) 

(i)

Rule : Transform into an integral of a rational function by the substitution e^ax = t.

Example 49 : Evaluate : 

Let I =  

Divide above and below by e^x then

Put 1 + e^x = t  ∴ e^x dx = - dt

then

= = - In t + c

= - In (1 + e^-x) + c

(ii) 

Rule : Express numerator as / (Denominator) + md/dx(Denominator) 

Find l and m by comparing the coefficients of ex and e^-x and split the integral into sum of two integrals as

= /x + m /n I Denominator I + c

Example 50 : Evaluate : 

Let I =

Express Numerator as

⇒ Numerator =/ (Denominator) + md/dx(Denominator)

or (4e^x + 6e^-x) = /(9e^x - 4e^-x) + m(9e^x + 4e^-x)

Comparing the coefficients of e^x and e^-x on both sides then 

4 = 9/ + 9m and 6 = - 4 / + 4m

After solving we get

then I =

=

=

The Integral of A Binomial Differential

A binomial differential is a differential of the form x^m (a + bxⁿ)^p dx where m, n, p ∈ Q and a, b are constants not equal to zero. The integral

is expressible in terms of elementary functions in the following three cases :

Case 1 : (i) If p ∈ N then expand by the formula of Newton Binomial. 

(ii) If p < 0, then we put x = t^k where k is the common denominator of the fractions m and n.

Case 2 : If  = Integer then we put a + bxⁿ = t^α where α is the denominator of the fraction p.

Example 51 : Evaluate :

Let I =

=

=

Put x² + 2x = t

∴

∴

=

DEFINITE INTEGRALS AND THEIR PROPERTIES

Definition : Let f be a function which is continuous on the closed interval [a, b]. The definite integral of f(x) from a to b is defined to be the limit

where

is a Riemann sum of f(x) on [a, b]

Newton-Leibnitz Formula 

Let f be a function of x defined in the closed interval [a, b] and the function F is an antiderivative of f on [a, b], such that F’(x) = f(x) for all x in the domain of f, then

= (F(b) + c) - (F(a) + c) 

= F(b) - F(b)

Note : In definite integrals, constant of integration is never present.

Rule for Finding Definite Integral

To evaluate the definite integral

First find out the indefinite integral of f(x) i.e., ∫f(x)dx, leaving the constant of integration c.

Let

=

To evaluate above the following cases may arise :

Case I : If f(x) is continuous at x = a and x = b :

∴  and

then

Case II : If f(x) is continuous at x = a and discontinuous at x = b :

∴

then

Case III : If f(x) is discontinuous at x = a and continuous at x = b :

∴

Case III : If f(x) is discontinuous at x = a and continuous at x = b :

∴

then

Case IV : If f(x) is discontinuous at x = c (a < c < b) : 

f(x) is discontinuous at x = c or f(x) has different values in (a, c) and (c, b) then

Note : If f(x) is not defined at x = a and x = b and defined in open interval (a, b) then  can be evaluated.

Example 51 : Evaluate :

= 244/5

Geometrical Interpretation of the Definite Integral 

First we construct the graph of the integrand y = f(x) then in the case of f(x) > 0 ∀ x ∈ [a, b], the integral J f(x) dx is numerically equal to the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x = b.

 dx is numerically equal to the area of curvilinear trapezoid bounded by the given curve, the straight lines x = a and x = b, and the x-axis.

In general : dx represents an algebraic sum of areas of the region bounded by the curve y = f(x), the x-axis and the ordinates x = a and

x = b. The areas above the x-axis enter into this sum with a positive sign, while those below the x-axis enter it with a negative sign.

∴

where A₁, A₂, A₃, A₄, A₅ are the areas of the shaded regions.

Example 52 : Evaluate : 

To find

when x < 0 

y = max{2 - x, 2, 1 + x} = 2 - x 

when x > 0 

y = max{2 - x, 2, 1 + x) = 2 

Plotting these curves on the XY plane

 = Area of shaded region

= Area of square ABCD + Area of ADEF 

= 2 x 2 + 1/2 x 1 x 1

= 4 + 1/2

= 9/2

Example 53 : Evaluate :

The graph of f(x) = min {I x I, I x - 1 |, | x + 1 |} is shown as in Fig..

Therefore  = Area of shaded region 

=

Example 54 : Evaluate :  where [x] denotes the greatest integer ≤ x. 

I =

=

=

= 1 + 1 + 3 

= 5

Definite Integral by standard methods of indefinite integral

(i) Transformation Method : To evaluate the integral dx we can transform f(x) into other function ∅(x) without any substitution then

and then  evaluate easily.

Example 56 : Evaluate :

Let I =

=

=

=

=

=

(ii) Substitution Method : By the substitution x = ∅(x) in the definite integral ∫ f(x)dx, the lower and the upper limits change.

When x = a then a = ∅(t) ⇒ t = ∅^-1(a) 

and when x = b then b = ∅(t) ⇒ t = ∅^-1 (b)

Hence the new lower and upper limits are ∅^-1(a) and ∅^-1(b) respectively.

Again put In

∴

⇒

∴

When t = 0 ⇒ z = 0 

t = ∝ ⇒ z = ∝

=

=

=

=

(iii) By Parts Method : Let u and v are the functions of x, u and v have continuous derivatives on the interval [a, b]. Then

where u' and v’ are the derivatives of u and v respectively.

Example 57 : Evaluate :

Integration by parts, Definite Integral

=

=

=

=

=

=

=

=

=

Properties of The Definite Integral

Some simple properties of definite integrals can be derived from the basic definition, or from the Fundamental Theorem of the Calculus.

(a)

If the upper and lower limits of the integral are the same, the integral is zero. This becomes obvious if we have a positive function and can interpret the integral in terms of ‘the area under a curve’.

(b) If a ≤ b ≤ c,

This says that the integral of a function over the union of two intervals is equal to the sum of the integrals over each of the intervals. The diagram opposite helps to make this clear if f(x) is a positive function.

(c)  for any constant c.

This tells us that we can move a constant past the integral sign but we can only do this with constants, never with variables!

(d)

That is, the integral of sum is equal to the sum of the integrals.

(e) If f(x) ≤ g(x) in [a, b] then

That is, integration preserves inequalities between functions. The diagram opposite explains this result if f(x) and g(x) are positive functions.

(f)

This tells us that the integral of a constant is equal to the product of the constant and the range of integration. It becomes obvious when we look at the diagram with c > 0, since the area represented by the integral is just a rectangle of height c and width b - a.

(g) We can combine (e) and (f) to give the result that, if M is any upper bound and m any lower bound for f(x) in the interval [a, b], so that m ≤ f(x) ≤ M, then

This, too becomes clear when f(x) is a positive function and we can interpret the integral as the area under the curve. 

(h) Finally we extend the definition of the definite integral slightly, to remove the restriction that the lower limit of the integral must be a smaller number than the upper limit. We do this by specifying that

For example,

Example 58 : Evaluate :

Let I =

Case I : When 0 < a < b

then

= I b I - | a |                                                                       -------------(1)

Case II : When a < 0 < b

then I =

= - (0 - a) + (b - 0) 

= b + a ...(2) 

= I b I - | a |                                                                       -------------(2)

Case III : When a < b < 0

then

= I b I - | a |                                                                       -------------(3)

It is clear from (1), (2) and (3) we get

Example 59 : Evaluate :  

Let I =

=                  

=

=

=

Some other properties

Property :

Proof : Analytical Method :

R.H.S. =

Put a - x = t  ∴dx = - dt

When x = 0 ⇒ t = a 

x = a ⇒ t = 0

∴ R.H.S. =

=

=

=

= L.H.S.

Removal of x : Let I =  where f(x) is a function of x whose integral is known and f(a - x) = f(x) then apply above property.

Example 60 : Evaluate :

Let I =

                                                              ...(1)

=

=                                                                ...(2)

Adding (1) and (2),

Hence I = π/4

Hence remember that :

(i)

(ii)

(iii)

(iv)

(v)

(vi) 

Example 61 : Evaluate :

Let I =

put t = 1 - x ⇒ dt = - dx

 at x = 0 ⇒ t = 1 

and x = 1 ⇒ t = 0

=

=

=

=

=

Example 62 : Evaluate : 

Let I =                                                           ...(1)

=

=                                                           ...(2)

Adding ( 1 ) and (2), we get

Put cos = t ∴ sin x dx = - dt

When x = 0 ⇒ t = 1 

x = π ⇒ t = -1

∴

=

=

Hence I = π²/4

Remember that :

(i)

(ii)

(iii) 

Property :

Note : In 95% cases a + b = π/4, π/2, π and in 5% cases a + b = 1, 2, 3, λ etc.

Example 63 : Evaluate : 

Let I =

=                                    ----(1)

=

=                                     ---(2)

adding (1) and (2), we get

= π/6

Hence I = π/12

Property :

Example 64 : Evaluate :

Let I =

=

=  

=

=

=

=

=

=

Property :

Example 65 : Evaluate :

Let I =                                                                    ----(1)

=

=                                                                           ----(2)

Adding (1) and (2),

=

=

=

Put 2x = t ⇒ dx = dt/2

When x = 0 ⇒ t = 0 

x = π/2 ⇒ t = π

∴

Here if f(t) = In sin t

f(π - t) = In sin (π - t) = In sin t = f(t)

then

=

∴

Hence remember that :

(i)

(ii)

(iii)

Example 66: Show that  = 

Let I =                                                                     ----(1)

=

=

=

Adding (1) and (2), we get

=

=

Put x = π/4 - t => dx = - dt

When x = 0 ⇒ t = π/4 

x = π/2 ⇒ t = -π/4

∴

=

=

=

Hence I =

Property : (Time Saving Property or Help line property)

and

Analytical Method :

Put x + c = t dx = dt

W hen x = a - c ⇒ t = a 

x = b - c ⇒ t = b

then

Also put x - c = t 

∴ dx = dt

When x = a + c ⇒ t = a 

x = b + c ⇒ t = b

then

Example 67 : Show that the sum of two integrals  is zero.

Let I =

= 

=

Put 3x = - t in second integral then 

3 dx = - dt

When x = - 1/3 ⇒ t = 1 

x = 0 ⇒ t = 0

then

=

=

Example 68 : If f(x) =  then prove that

R.H.S. =

=

=

=

=

=

put at at

t - a = u ⇒ dt = du 

t = 1 + a we get u = 1 

t = x + a we get u = x

=

=

= L.H.S.

Property : (Time Saving property or Help line property)

and

Example 69 : Evaluate :

Let 2πx = t ⇒ 2π dx = dt ⇒ dx = dt/2

if x = 0 ⇒ t = 0 ; x = 1 ⇒ t = 2π

⇒

=

=

=

=

Example 70 : Show that

Let I =  (here x is constant)

put  ∴

when z = 0 ⇒ t = - x 

z = x ⇒ t = x

∴

=

=

=  (∴ Function is even)

=

OR

=

=

=

=

put z - x/2 = t ⇒ dz = dt 

at z = 0 we get t = -x/2 

at z = x we get t = x/2

=

put t = u/2 ⇒ dt = du/2

at t = -x/2 we get u = -x 

at t = x/2 we get u = x

=

=  (∴ Function is even)

=

= R.H.S.

Property : (Reflection Property)

Proof. Analytical Method :

Let I =                                                                     ----(1)

Put - x = t  ∴dx = - dt

when x = - b ⇒ t = b 

x = - a ⇒ t = a

then

=

=                                                                                 ----(2)

From (1) and (2), we get

Graphical Method 

The curve y = f(-x) is obtained with the help of y = f(x), v y = f(-x) is the mirror image of y = f(x) in the y-axis. It is clear from the Fig..

Area of ABCDA = Area of BA’C’D’A 

then

Example 71 : Evaluate :

Let I =

=

=                                                                      ----(1)

and

=                                                                       ----(2)

Adding (1) & (2) we get

∴ I = π/12

Property : If f(x) is discontinuous at x = a then 

Proof : R.H .S . =

=

=

=

=

Example 72 : Evaluate :

Let I =

Integrand is discontinuous at x = π/2 then

 

=

=

=

=

Put  then cos x dx =

∴

=

=

=

Property :

Example 73 : Given , find the value of  in terms of α.

Let I =

Where a = 4π - 2 , b = 4π, f( t) =

according to property

I =

=  

=    (substitute t = 1- t)

=

=      

=

= - α                     (given)

Property :

where T is the period of the function and n ∈ I 

i.e., f(x + T) = f(x)

Example 74 : Evaluate :

Let I =   

=

=

=  ( ∴ I cos x I is periodic with period π )

=

=  

=

=

= 200√2

Property : 

where T is the period of the function and m, n ∈ I 

i.e., f(x + T) = f(x)

Proof. L.H.S. =

=     (∴ f(x) is periodic)

=

= R.H.S.

Example 75 : Evaluate :

Let I =

=

=       ( ∴I sin x I is periodic with period π )

=

=

= 200 (1 - (-1)) = 400

Property :

where T is the period of the function and n ∈ I 

i.e., f(x + T) = f(x)

Proof. L.H.S. =

=     ( ∴ f(x + nT) = f(x))

= R.H.S.

Example 76 : Show that  where n ∈ N and 0 ≤ V ≤ π.

Let I =

=

=    (∴ I sin x I is periodic with period π) 

=          ( ∴ 0 ≤ V ≤ n )

=

=  n(1 - ( - 1 ) ) + ( - cos V + 1)

= (2n + 1) - cos V

Example 77 : Find the value of

I =

=

=

=

Property : Leibnitz’s Rule :

(i) If u(x) and v(x) are differentiable functions of x whose values lie in [a, b] and f(x, t) is continuous then

Note : Where d/dx f(x, t)

represents the differentiation with respect to x treating t as  constant.

(ii) If f is continuous function on [a, b] and u(x) and v(x) are differentiable functions of x whose values lie in [a, b] then

Example 78 : If  then evaluate dy/dx.

We have

Differentiating both sides w.r.t. x then

⇒

⇒

∴

Property :

 s independent of a

where T is the period of the function 

i.e., f(x + T) = f(x)

Proof. Let I =

=

= f(a + T) . 1 - f(a) . 1 

= f(a + T) - f(a) 

= 0                                     { ∴ f(a + T) = f(a )}

Thus, I = c (independent of a)

Hence  s independent of a

Example 79 : Evaluate :

∴ sin⁴ x + cos⁴ x is periodic with period π/2 then

=

=

=

=

=

=

=

=

Property : Suppose that f(x, a) and f_α’(x, a) are continuous functions when c ≤ α ≤ d and a ≤ x ≤ b then

the derivative of 1(α) w.r.t. α and f_α’(x, α) is the derivative of f(x, α) treating x as constant.

Example 80: Evaluate :

Let l(a) =  

Where

then

According to property -  where I‘(a) is the derivative of l(a) 

∴

=

Integrating both sides w.r.t. a then 

l(a) = In (a + 1) + c 

for a = 0, l(0) = In 1 + c 

0 = 0 + c ⇒ c = 0 

by equation (1) 

∵ I(a) = ln(a + 1)

Property :

If f(t) is an odd function then ∅(x) =  is an even function. 

Proof. Since ∅(x) =

∴

=

=

=

=

=

= ∅(x)

Hence ∅(-x) = ∅(x)

Example 81: Prove that F(x) =  is an even function.

Let

∴

=

= -f(t)

⇒ f(-t) = -f(t)    ∴ f(t) is an odd function.

Hence F(x) =  is an even function

Property :

If f(t) is an even function then ∅(x) =  s an odd function.

Proof. The proof is similar as above. Follow the same steps considering f(t) to be an even function.

 Example 82 : Prove that F(x) =  dt is an odd function.

Let

∴

=

=

=

=

=

= f(t)

⇒ f(-t) = f(t)                             ∴f(t) is an even function.

Hence  is an odd function.

Important Note : If f(t) an even function then  is not necessarily an odd function for a ≠ 0. It will be an odd function if

Proof. Suppose  

⇒

⇒

⇒

⇒

⇒

Property : If on an interval [a, b] (a < b), and the functions f(x) and g(x) satisfy the condition f(x) ≥ g(x) then

In Particular :

If f(x) ≥ 0 then

Proof : Graphical Method :

As we know that  denotes the area under f(x) from a to b and if f(x) ≥ g(x) ∀ x ∈ (a, b) then clearly area under f(x) from a to b which can be seen by given fig.

It is clear from the Fig. Area of curvilinear trapezoid aBCb ≥ Area of curvilinear trapezoid aADb then  and if

Example 83 : Prove that :

∴ 0 < x < 1

then x > x² => - x < - x²

e^x < e^x2

then

Property : If at every point x of an interval [a, b] the inequalities  are fulfilled then

Proof. Graphical Method : 

It is clear from the Fig.

Area of curvilinear trapezoid aAFb 

≤ Area of curvilinear trapezoid aBEb 

≤ Area of curvilinear trapezoid aCDb

i.e.,

Example 84: Prove that :

Since  then 

Hence

Property : If m is the least value (global minimum) and M is the greatest value (global maximum) of the function f(x) on the interval [a, b] (estimation of an integral). Then

Proof : Analytical Method :

It is given that m ≤ f(x) ≤ M

then

⇒

Graphical Method : It is clear from the Fig..

Area of abDA ≤  ≤ Area of aBCb

i.e.,

Example 85: Prove that

Let

∴

for critical points f’(x) = 0 

∴ x = 9 and x = 1 

∵ x ∈ [0, 2]) so we take x = 1.

then 

f(0) = 5/9 

f(1) = 1/2 

f(2) = 3/5 

∴ The greatest and the least values of the integral in the interval [0, 2] are respectively, equal to f(2) = 3/5 and f(1) = 1/2. Hence

or

Property : The following inequality is valid :

    (a ≤ b)

or if a is not necessarily less than b, then

if f and I f I are integrable on [a, b].

Proof. Obviously,

⇒                 (a < b)

or

or

If

=

Hence

Example 86 : Estimate the absolute value of the integral

Since I sin x I ≤ 1 for x ≥ 10 then

                             ---(1)≈

but 10 ≤  x ≤ 19 

1 + x⁸ ≥ x⁸ ≥ 10⁸

⇒

or

from (1) & (2) we get

∴

= (19 - 10) x 10^-8 

= 9x 10^-8 

= (10 - 1) x 10^-8 

= 10^-7 - 10^-8 < 10^-7

Hence

∴ The true value of the integral ≈ 10^-8

Property : 

If f²(x) and g²(x) are integrable on the interval [a, b], the Schwarz-Bunyakovsky inequality takes place :

Proof. Let F(x) = (f(x) - λg(x)}² > 0

∴

⇒

⇒

∴ Discriminant is one positive i.e., B² - 4AC ≤ 0

⇒

Hence

Example 87: Prove that

=

=

=

Hence

Property : If f(x) is continuous on [a, b] then there exists a point c ∈ (a, b) such that 

Then number  is called the Mean value of the function f(x) on the interval [a, b].

Proof : Analytical Method :

For a < b. If m & M are smallest and greatest values of f(x) on [a, b]

then

or

Since f(x) is continuous on [a, b], It takes on all intermediate values between m and M. Therefore, some values f(c) (a ≤ f(c) ≤ b), we will have

or

Graphical Method :

It is clear from the Fig.

Area of aABb = Area of aDEb

i.e.,  or

Note : Used of this property also in physics if we evaluate average of physical quantities.

i.e., Average speed =

Example 88: Find the mean value of x(a - x) over the range (0, a).

Mean value =

=

=

=

The Root Mean Square Value (RMSV) of a function y = f(x) the range (a, b) is

Example 89: Find RMSV of In x over the range x = 1 to x = e

Here y = In x

                                                 ...(1)

Now

=

=

=

∴

= (e - 2e + 2e) - (0 - 0 + 2) 

= e - 2

From (1), we get

Property :

Example 90: Evaluate :  where [.] denotes the greatest integral function

Here f(x) = x² + 1 and a(x) = [x] 

df(x) = 2xdx

According to property,

∴

⇒

=

=

= 130 - 70 = 60

Property : 

(i) If the function f(x) increases and has a concave graph in the interval [a, b], then

(ii) If the function f(x) increases and has a convex graph in the interval [a, b] then

Example 91: Prove that :

Let f(x) =  

∴

∴ graph is concave up.

∴

⇒

Property : Improper Integrals :

(i)

(ii)

(iii)

Example 92: Prove that

Put In

or    ⇒

or     ∴

∴

when x → 0, t → 0 and x → ∞, t → ∞

L.H.S. =

=

=

=

=

=

Definite Integral as the limit of a sum (Integration by First Principle Rule)

Let f(x) be a single valued continuous function in the interval (a, b), (a < b) and if the interval (a, b) be divided into n equal parts, each of width h, we have b - a = nh

when n → ∞ then h→ 0

∴

=

=

=

Hence

Some Important Results to Remember

(i)

(ii)

(iii)

(iv)

(v)

(vi) In G.P., sum of n terms , S_n =

(vii) =

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv) 

Example 93: Prove that  by first principle rule. 

Here f(x) = x², nh = b - a from the definition,

=

=

=

=

=

=

=

Example 94: From the definition of a definite integral as the limit of sum evaluate

Here f(x) = e^x, nh = b - a from the definition

=

=

=

=

=

=

= e^b - e^a

Summation of series by Integration

Let f(x) be a continuous function defined on the closed interval [a, b], then 

Rule : Here we proceed as follows :

(i) Express the given series in the form

(ii) Then the limit is its sum when n → ∞

=

Replace r/n by x, 1/n by dx and  sign of ∫.

(iii) The lower and upper limits of integration will be the values of r/n for the first and last term (or the limit of these values) respectively.

when

Note : 

Example 95: Find the limit, when n → ∞ of

Let P =

=

=

=

let r/n = x 

at r = 1 and as n → ∞, we get x → 0 

at r = n and as n → ∞, we get x →1 

Put x = t²   ∴ dx = 2t dt

when x = 0 ⇒ t = 0 

x = 1 ⇒ t = 1

∴

=

Hence P = π/2.

Example 96: Find the limit, when n → ∞ of

Let P =

P =

=

=

=

Put 3√x + 4 = t

∴

when x = 0 ⇒ t = 4 

x = 1 ⇒ t = 7

∴

=

=

Hence P = 1/14

Gamma Function

The definite integral  is called the Second Eulerian integral and is denoted by the symbol Гn (read as Gamma n).

∴

⇒

This formula is known as recurrence formula for Gamma function.

Important Formula :

where m > -1 and n > -1 

(This formula is applicable only when the limit is 0 to π/2).

Note that : 

Walli’s formula

[An easy way to evaluate  where m, n ∈ I₊]

we have

where p is π/2 if m and n are both even, otherwise p = 1. In last factor in each of the three products is either 1 or 2. In case of m or n is 1, we simply write 1 as the only factor to replace its product. If any of m or n is zero provided we put 1 as the only factor in its product and we regard 0 as even.

Problem on Greatest Integral Function

Example 96: Evaluate : w here [.] denotes the greatest integer function .

Analytical Method :

Value of x² at x = 0 is 0² = 0 and at x = 1 -5 is (1 -5)² = 2-25 

∴ integers between 0 to 2-25 are 1 , 2

then x² = 1 and x² = 2 x = 1 and x = √2

=

=

=  (2 - √2)

Graphical Method :

 = = Area of shaded region

=

=  (2 - √2)

Example 97: Prove that :  where [.] denotes the greatest integer function.

Analytical Method :

Let x = n + f ∀ n ∈ I and 0 ≤ f < 1 

∴ [x] = n                                                                 ----(1)

=

= 1 + 2 + 3 + ... + (n - 1) + nf

=

=                                          [from (1)}

Graphical Method :

Let x = n + f ∀ n ∈ I and 0 ≤ f < 1 

∴ [x] = n                                                                 ----(1)

 =  Area of shaded region 

=  0 + 1 + 2 + ... + (n - 1) + (x - n) n

=

=

Example 98:  Evaluate :  where [.] denotes the greatest integer function.

Analytical Method :

∵ 0 ≤ x < π 

0 ≤ 2 sin x ≤ 2

Now when 2 sin x = 1⇒ sin x = 1/2

x = π/6, π/6

and If 0 ≤ 2 sinx < 1 , 0 < x < π/6 and

∴ [2 sin x] = 0 and If 1 ≤ 2 sin x < 2

∴

Now

=

Graphical Method:

 =  Area of shaded region

=

= 2π/3