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Properties of Definite Integrals - III

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Properties of Definite Integrals - III - Lesson Summary

Property 6:

02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a - x) dx

Proof:

02a f(x) dx

By a property on definite integrals,

if a < c < b , then ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx

02a f(x) dx = ∫0a f(x) dx + ∫a2a f(x) dx -----(1)

Let's evaluate ∫02a f(x) dx

Let t = 2a - x ⇒ x = 2a - t

dx = 0 - dt ⇒ dx = -dt

When x = a, t = 2a - a = a

When x = 2a, t = 2a - 2a = 0

02a f(x) dx = ∫a0 f(2a - t) x (-dt)

 = - ∫a0 f(2a - t) dt

By a property of definite integrals,

ab f(x) dx = - ∫ba f(x) dx

   = ∫0a f(2a - t) dt

By a property of definite integrals,

ab f(x) dx = ∫ab f(t) dt

      =  ∫0a f(2a - x) dx

∴ ∫02a f(x) dx = ∫0a f(2a - x) dx ----(2)

On substituting equation 2, in equation 1, we get

02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a - x) dx

Example:

Evaluate ∫0π  (x sin x)/(1 + cos2x) dx.

Let I = ∫0π  (x sin x)/(1 + cos2x) dx

= ∫02π/2  (x sin x)/(1 + cos2x) dx

02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a - x) dx

= ∫02×π/2  (x sin x)/(1 + cos2x) dx

= = ∫0π/2  (x sin x)/(1 + cos2x) dx +  ∫0π/2  ((π- x) sin (π-x))/(1 + cos2(π-x)) dx

 = ∫0π/2  (x sin x)/(1 + cos2x) dx +  ∫0π/2  ((π- x) sin x)/(1 + (- cos x )2) dx

 = ∫0π/2  (x sin x)/(1 + cos2x) dx +  ∫0π/2  (π sin x - x sin x)/(1 + cos2 x) dx

= ∫0π/2  (x sin x + π sin x - x sin x)/(1 + cos2x) dx

= ∫0π/2  (π sin x)/(1 + cos2x) dx

= π ∫0π/2  sin x/(1 + cos2x) dx -----(1)

Let cos x = t

⇒ - sin x dx = dt

⇒ sin x dx = - dt

When x = π/2 , t = cos π/2 = 0

When x = 0 , t = cos 0 = 1

= π ∫10 -1/(1+t2) dt

= -π ∫10 dt/(1+t2)

ab f(x) dx = - ∫ba f(x) dx

 -π ∫10 dt/(1+t2) = = π ∫01 dt/(1+t2)

∫ 1/(1+t2) dt = tan-1t + C

π ∫ dt/(1+t2) = π [tan-1t]01

                  = π [tan-11 - tan-10]

                  = π(π/4 - 0)

                   = π2/4


Property 7:

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

Proof:

02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a - x) dx----(1)

If f(2a - x) = f(x),   then from (1) , we have

02a f(x) dx = ∫0a f(x) dx + ∫0a f(x) dx

                = 2 . ∫0a f(x) dx

If f(2a - x) = - f(x), then from (1), we have

02a f(x) dx = ∫0a f(x) dx - ∫0a f(x) dx

                = 0

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

Example:

Evaluate ∫0π tan x/(sec x + cos x) dx

Let f(x) = tan x/(sec x + cos x)  and 2a = π

Let's find out f(2a - x) = f(π-x) = tan (π-x)/(sec (π-x) + cos (π-x))

                               = - tanx/(-secx-cosx)

                               = tanx/(secx + cosx)

                               = f(x)

By a property on definite integrals,

02a f(x) dx = 2. ∫0a f(x) dx if f(2a - x) = f(x)

∴ ∫0π tanx/(secx + cosx) dx = 2 ∫0π/2 tanx/(secx + cosx) dx

                                        = 2 ∫0π/2 (sin x/cox x)/(1/cos x + cosx) dx

                                         = 2 ∫0π/2 sin x/(1 + cos2x) dx

Let cos x = t

On differentiating both sides, we get

- sin x dx = dt

sin x dx = -dt

When x = 0 , t = cos 0 = 1

When x = π/2; t = cos π/2 = 0

Therefore, 2 ∫0π/2 sin x/(1 + cos2x) dx = 2.∫10 -dt/(1+t2)

ab f(x) dx = - ∫ba f(x) dx

2 .∫10 -dt/(1+t2) = 2 .∫01 1/(1+t2) dt

                       = 2 [tan-1 t]01

                        = 2[tan-1(1) - tan-1(0)]

                        = 2[π/4 - 0]

                        = π/2

0π tanx/(sec x + cos x) dx = = π/2



Property 8:

i. ∫-aa f(x) dx = 2 ∫0a f(x) dx,

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

ii. ∫-aa f(x) dx = 0

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

Proof:

i. LHS = ∫-aa f(x) dx

If a < c < b, then

ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx

We have - a < 0 < a and

-aa f(x) dx = ∫-a0 f(x) dx + ∫0a f(x) dx  ------(1)

-a0 f(x) dx

Let x = -t

dx = -dt

When x = -a, then -t = a ⇒ t = a

When x = 0 , then -t = 0 ⇒ t = 0

-a0 f(x) dx = ∫-a0 f(-t) (-dt)

 = - ∫a0 f(-t) dt

 = ∫0a f(-t) dt      ['.' ∫ab f(x) dx = - ∫ba f(x) dx]

-a0 f(x) dx = ∫0a f(-t) dt ----(2)

However, by a property of definite integrals,

ab f(x) dx = ∫ab f(t) dt

-a0 f(x) dx = ∫0a f(-x) dx -----(3)

  ∫-aa f(x) dx = ∫0a f(-x) dx + ∫0a f(x) dx -----(4)

Case - 1:If the function is an even function, i.e.

f(-x) = f(x):

-aa f(x) dx = ∫0a f(-x) dx + ∫0a f(x) dx

⇒ ∫-aa f(x) dx = ∫0a f(x) dx + ∫0a f(x) dx

⇒ ∫-aa f(x) dx = 2 ∫0a f(x) dx

Case - 2: If the function is an odd function, i.e.

f(-x) = - f(x):

⇒ ∫-aa f(x) dx = ∫0a f(-x) dx + ∫0a f(x) dx

⇒ ∫-aa f(x) dx = ∫0a -f(x) dx + ∫0a f(x) dx

⇒ ∫-aa f(x) dx = - ∫0a f(x) dx + ∫0a f(x) dx

⇒ ∫-aa f(x) dx = 0

i. ∫-aa f(x) dx = 2. ∫0a f(x) dx,

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

ii. ∫-aa f(x) dx =  0

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

Example:

Evaluate ∫log½log2 sin ((ex-1)/(ex+1))dx

log½log2 sin ((ex-1)/(ex+1))dx = ∫-log2log2 sin ((ex-1)/(ex+1))dx

Let f(x) = sin ((ex-1)/(ex+1))

Now, f(-x) = sin ((e-x-1)/(e-x+1))

 = sin ((e1/x-1)/(e1/x+1))

= sin ((1 - ex)/(1+ex))

= sin (-(ex-1)/(1+ex))

= - sin ((ex-1)/(ex+1)) ['.' sin(-x) = - sin x]

= - f(x)

⇒ f(-x) = - f(x)

∴ f(x) is an odd function

-aa f(x) dx =  0

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

log½log2 sin ((ex-1)/(ex+1))dx = 0

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