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# Properties of definite integrals - II

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#### Properties of definite integrals - II - Lesson Summary

Property 4:

âˆ«ab f(x) dx = âˆ«ab f(a + b - x) dx

Proof:

Let a + b - x = t â‡’ x = a + b - t

dx = - dt

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

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

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

= - âˆ«ab f(a + b - t) dt

By a property of definite integrals, we have

âˆ«ab f(x) dx = - âˆ«ba f(x) dx

= âˆ«ab f(a + b - t) dt

By a property of definite integrals, we have

âˆ«ab f(x) dx = âˆ«ab f(t) dt

= âˆ«ab f(a + b - x) dx

âˆ«ab f(x) dx = âˆ«ab f(a + b - x) dx

Ex:

Evaluate âˆ«-Ï€/2Ï€/2 cos x /(1 + ex) dx

Let I = âˆ«-Ï€/2Ï€/2 cos x /(1 + ex) dx----- (1)

By a property of definite integrals, we have

âˆ«ab f(x) dx = âˆ«ab f(a + b - x) dx

â‡’ I = âˆ«-Ï€/2Ï€/2 cos (-Ï€/2 + Ï€/2 - x) /(1 + e-Ï€/2 + Ï€/2 - x) dx

I = âˆ«-Ï€/2Ï€/2 cos (-x) /(1 + e-x) dx

I = âˆ«-Ï€/2Ï€/2 cos x /(1 + e-x) dx

I = âˆ«-Ï€/2Ï€/2 cos x /(1 + 1/ex) dx

= âˆ«-Ï€/2Ï€/2 ex cos x /(ex + 1) dx -----(2)

Adding (1) and (2), we get

2I = âˆ«-Ï€/2Ï€/2 cos x /(1 + 1/ex) dx + âˆ«-Ï€/2Ï€/2 ex cos x /(ex + 1) dx

= âˆ«-Ï€/2Ï€/2 ( cos x /(1 + 1/ex)  + (ex cos x) /(ex + 1)) dx

âˆ«-Ï€/2Ï€/2 (cos x + ex cos x) /(1 + ex ) dx

= âˆ«-Ï€/2Ï€/2 ( 1 + ex ) cos x /(1 + ex ) dx

= âˆ«-Ï€/2Ï€/2 cos x dx

= [sin x]-Ï€/2Ï€/2

= sin(Ï€/2) - sin (-Ï€/2)

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

= 1 + 1 = 2

â‡’ 2I = 2  â‡’ I = 1

âˆ´ âˆ«-Ï€/2Ï€/2 cos x /(1 + 1/ex) dx = 1

Property 5:

âˆ«0a f(x) dx = âˆ«0a f(a - x) dx

Proof:

Let a - x = t â‡’ x = a - t

dx = - dt

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

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

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

= - âˆ«ao f(a - t) dt

By a property on definite integrals, we have

âˆ«ab f(x) dx = - âˆ«ba f(x) dx

= âˆ«0a f(a - t) dt

By a property of definite integrals, we have

âˆ«ab f(x) dx = âˆ«ab f(t) dt

= âˆ«oa f(a - x) dx

âˆ«0a f(x) dx = âˆ«0a f(a - x) dx

Ex:

Evaluate âˆ«0Ï€/2 (cos x - sin x)/(1 + cos x sin x) dx

Let I = âˆ«0Ï€/2 (cos x - sin x)/(1 + cos x sin x) dx -----(1)

â‡’ I = âˆ«0Ï€/2 (cos (Ï€/2 - x) - sin (Ï€/2 - x))/(1 + cos (Ï€/2 - x) sin (Ï€/2 - x)) dx

I = âˆ«0Ï€/2 (sin x - cos x)/(1 + sin x cos x) dx -----(2)

Adding (1) and (2), we get

2I = âˆ«0Ï€/2 (cos x - sin x)/(1 + cos x sin x) dx + âˆ«0Ï€/2 (sin x - cos x)/(1 + sin x cos x) dx

= âˆ«0Ï€/2 (cos x - sin x + sin x - cos x)/(1 + cos x sin x) dx

= 0

â‡’ 2I = 0

â‡’ I = 0

Hence, âˆ«0Ï€/2 (cos x - sin x)/(1 + cos x sin x) dx = 0