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Properties of definite integrals - II

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Properties of definite integrals - II - Lesson Summary

Property 4:

ab f(x) dx = ∫ab f(a + b - x) dx

Proof:

Let a + b - x = t ⇒ x = a + b - t

dx = - dt

When x = a, t = a + b - a = b

When x = b, t = a + b - b = a

Therefore, ∫ab f(x) dx = ∫ab f(a + b - t) (-dt)

     = - ∫ab f(a + b - t) dt

By a property of definite integrals, we have

ab f(x) dx = - ∫ba f(x) dx

= ∫ab f(a + b - t) dt

By a property of definite integrals, we have

ab f(x) dx = ∫ab f(t) dt

= ∫ab f(a + b - x) dx

ab f(x) dx = ∫ab f(a + b - x) dx

Ex:

Evaluate ∫-π/2π/2 cos x /(1 + ex) dx

Let I = ∫-π/2π/2 cos x /(1 + ex) dx----- (1)

By a property of definite integrals, we have

ab f(x) dx = ∫ab f(a + b - x) dx

⇒ I = ∫-π/2π/2 cos (-π/2 + π/2 - x) /(1 + e-π/2 + π/2 - x) dx

I = ∫-π/2π/2 cos (-x) /(1 + e-x) dx

I = ∫-π/2π/2 cos x /(1 + e-x) dx

I = ∫-π/2π/2 cos x /(1 + 1/ex) dx

= ∫-π/2π/2 ex cos x /(ex + 1) dx -----(2)

Adding (1) and (2), we get

2I = ∫-π/2π/2 cos x /(1 + 1/ex) dx + ∫-π/2π/2 ex cos x /(ex + 1) dx

= ∫-π/2π/2 ( cos x /(1 + 1/ex)  + (ex cos x) /(ex + 1)) dx

-π/2π/2 (cos x + ex cos x) /(1 + ex ) dx

= ∫-π/2π/2 ( 1 + ex ) cos x /(1 + ex ) dx

= ∫-π/2π/2 cos x dx

= [sin x]-π/2π/2

= sin(π/2) - sin (-π/2)

= sin(π/2) + sin (π/2)         ['.' sin (-π/2) = - sin(π/2)]

= 1 + 1 = 2

⇒ 2I = 2  ⇒ I = 1

∴ ∫-π/2π/2 cos x /(1 + 1/ex) dx = 1


Property 5:

0a f(x) dx = ∫0a f(a - x) dx

Proof:

Let a - x = t ⇒ x = a - t

dx = - dt

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

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

Therefore, ∫0a f(x) dx = ∫ao f(a - t) (-dt)

= - ∫ao f(a - t) dt

By a property on definite integrals, we have

ab f(x) dx = - ∫ba f(x) dx

 = ∫0a f(a - t) dt

By a property of definite integrals, we have

ab f(x) dx = ∫ab f(t) dt

= ∫oa f(a - x) dx

0a f(x) dx = ∫0a f(a - x) dx

Ex:

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

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

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

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

Adding (1) and (2), we get

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

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

    = 0

 ⇒ 2I = 0

 ⇒ I = 0

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

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