Properties of definite integrals - I

Properties of definite integrals:

1. ab f(x) dx = ∫ab f(t) dt

2. abf(x) dx = - ∫abf(x) dx

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

Property 1:

ab f(x) dx = ∫ab f(t) dt

Ex:

Evaluate 01 1/√(1-x2) dx

Claim:  01 1/√(1-x2) dx = ∫01 1/√(1-t2) dt

We know that  ∫ 1/√(1-x2) dx = sin-1 x + C

⇒∫01 1/√(1-x2) dx = [sin-1 x]01

= sin-1 1 - sin-1 0

= π/2 - 0

= π/2

Hence, 01 1/√(1-x2) dx = π/2 -----(1)

01 1/√(1-t2) dt

We know that  01 1/√(1-t2) dt = sin-1 t + C

⇒ ∫01 1/√(1-t2) dt = [sin-1 t]01

= sin-1 1 - sin-1 0

π/2 - 0 = π/2

Hence, ∫01 1/√(1-t2) dt = π/2 ------(2)

∴ ∫01 1/√(1-x2) dx = ∫01 1/√(1-t2) dt


Property 2:

ab f(x) dx = - ∫ba f(x) dx

Also, if b = a, then ab f(x) dx = 0

Proof:

Let ∫ f(x) dx = G(x)


Second fundamental theorem on integral calculus:

If f is a continuous function defined on [a, b] and

∫ f(x) dx = G(x), then

ab f(x) dx = G(x)ab = G(b) - G(a)

ab f(x) dx = G(x)ab

    = G(b) - G(a)

    = - [G(a) - G(b)]

    = - [G(x)]ab

    = - ∫ba f(x) dx

ab f(x) dx = - ∫ba f(x) dx

If b = a, then ∫ba f(x) dx = ∫aa f(x) dx

∫ f(x) dx = G(x) + C

aa f(x) dx = G(x)aa = G(a) - G(a) = 0

Example:

I = ∫0π  sin x/(1+cos2x) dx

cos x = t

⇒ - sin x dx = dt

⇒ sin x dx = - dt

When x = 0,⇒ t = cos 0 = 1

When x = π , ⇒ t = cos π = -1

0π  sin x/(1+cos2x) dx = ∫1-1  -dt/(1+t2)

By property on integral calculus, we have

ab f(x) dx = - ∫ba f(x) dx

1-1  -dt/(1+t2) = - ∫-11  -dt/(1+t2)

-11  dt/(1+t2)

= (tan-1 t)-11             ('.'  ∫  dt/(1+t2) = tan-1 t)

= (tan-1 1 - tan-1 (-1))

= tan-1 1 + tan-1 1

= π/4 + π/4

= 2π/4

= π/2

Hence, 0π  sin x/(1+cos2x) dx = π/2



Property 3:

If a < c < b  then

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

Proof:

Let G(x) = ∫ f(x) dx

ab f(x) dx = G(b) - G(a) ….(i)

ac f(x) dx = G(c) - G(a)…..(ii)

cb f(x) dx = G(b) - G(c) ….(iii)

(ii) + (iii) =

ac f(x) dx + ∫cb f(x) dx = G(c) - G(a) + G(b) - G(c)

⇒ ∫ac f(x) dx + ∫ac f(x) dx = G(b) - G(a)

= ∫ab f(x) dx

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

Ex:

Evaluate 04 |2 - x| dx

Let I = ∫04 |2 - x| dx

We know that

If a < c < b, then

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

So, ∫04 |2 - x| dx = ∫02 |2 - x| dx + ∫24 |2 - x| dx

= ∫02 |2 - x| dx + ∫24 |2 - x| dx

= ∫02 2 dx - ∫02 x dx +  ∫24 x dx - ∫24 2 dx

= 2[x]02 - [x2/2]02 - [x2/2]24 - 2[x]24

= 2(2 - 0) - 1/2(22 - 0) + 1/2 (42 - 22) - 2(4 - 2)

= 4 - 1/2.(4) + 1/2(16 - 4) - 2(2)

= 4 - 2 + 6 - 4

= 4

Hence, 04 |2 - x| dx = 4

Summary

Properties of definite integrals:

1. ab f(x) dx = ∫ab f(t) dt

2. abf(x) dx = - ∫abf(x) dx

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

Property 1:

ab f(x) dx = ∫ab f(t) dt

Ex:

Evaluate 01 1/√(1-x2) dx

Claim:  01 1/√(1-x2) dx = ∫01 1/√(1-t2) dt

We know that  ∫ 1/√(1-x2) dx = sin-1 x + C

⇒∫01 1/√(1-x2) dx = [sin-1 x]01

= sin-1 1 - sin-1 0

= π/2 - 0

= π/2

Hence, 01 1/√(1-x2) dx = π/2 -----(1)

01 1/√(1-t2) dt

We know that  01 1/√(1-t2) dt = sin-1 t + C

⇒ ∫01 1/√(1-t2) dt = [sin-1 t]01

= sin-1 1 - sin-1 0

π/2 - 0 = π/2

Hence, ∫01 1/√(1-t2) dt = π/2 ------(2)

∴ ∫01 1/√(1-x2) dx = ∫01 1/√(1-t2) dt


Property 2:

ab f(x) dx = - ∫ba f(x) dx

Also, if b = a, then ab f(x) dx = 0

Proof:

Let ∫ f(x) dx = G(x)


Second fundamental theorem on integral calculus:

If f is a continuous function defined on [a, b] and

∫ f(x) dx = G(x), then

ab f(x) dx = G(x)ab = G(b) - G(a)

ab f(x) dx = G(x)ab

    = G(b) - G(a)

    = - [G(a) - G(b)]

    = - [G(x)]ab

    = - ∫ba f(x) dx

ab f(x) dx = - ∫ba f(x) dx

If b = a, then ∫ba f(x) dx = ∫aa f(x) dx

∫ f(x) dx = G(x) + C

aa f(x) dx = G(x)aa = G(a) - G(a) = 0

Example:

I = ∫0π  sin x/(1+cos2x) dx

cos x = t

⇒ - sin x dx = dt

⇒ sin x dx = - dt

When x = 0,⇒ t = cos 0 = 1

When x = π , ⇒ t = cos π = -1

0π  sin x/(1+cos2x) dx = ∫1-1  -dt/(1+t2)

By property on integral calculus, we have

ab f(x) dx = - ∫ba f(x) dx

1-1  -dt/(1+t2) = - ∫-11  -dt/(1+t2)

-11  dt/(1+t2)

= (tan-1 t)-11             ('.'  ∫  dt/(1+t2) = tan-1 t)

= (tan-1 1 - tan-1 (-1))

= tan-1 1 + tan-1 1

= π/4 + π/4

= 2π/4

= π/2

Hence, 0π  sin x/(1+cos2x) dx = π/2



Property 3:

If a < c < b  then

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

Proof:

Let G(x) = ∫ f(x) dx

ab f(x) dx = G(b) - G(a) ….(i)

ac f(x) dx = G(c) - G(a)…..(ii)

cb f(x) dx = G(b) - G(c) ….(iii)

(ii) + (iii) =

ac f(x) dx + ∫cb f(x) dx = G(c) - G(a) + G(b) - G(c)

⇒ ∫ac f(x) dx + ∫ac f(x) dx = G(b) - G(a)

= ∫ab f(x) dx

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

Ex:

Evaluate 04 |2 - x| dx

Let I = ∫04 |2 - x| dx

We know that

If a < c < b, then

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

So, ∫04 |2 - x| dx = ∫02 |2 - x| dx + ∫24 |2 - x| dx

= ∫02 |2 - x| dx + ∫24 |2 - x| dx

= ∫02 2 dx - ∫02 x dx +  ∫24 x dx - ∫24 2 dx

= 2[x]02 - [x2/2]02 - [x2/2]24 - 2[x]24

= 2(2 - 0) - 1/2(22 - 0) + 1/2 (42 - 22) - 2(4 - 2)

= 4 - 1/2.(4) + 1/2(16 - 4) - 2(2)

= 4 - 2 + 6 - 4

= 4

Hence, 04 |2 - x| dx = 4

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