Notes On Properties of definite integrals - I - CBSE Class 12 Maths
Properties of definite integrals: 1. âˆ«ab f(x) dx = âˆ«ab f(t) dt 2. âˆ«abf(x) dx = - âˆ«abf(x) dx 3. if a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cbf(x) dx Property 1: âˆ«ab f(x) dx = âˆ«ab f(t) dt Ex: Evaluate âˆ«01 1/âˆš(1-x2) dx Claim:  âˆ«01 1/âˆš(1-x2) dx = âˆ«01 1/âˆš(1-t2) dt We know that  âˆ« 1/âˆš(1-x2) dx = sin-1 x + C â‡’âˆ«01 1/âˆš(1-x2) dx = [sin-1 x]01 = sin-1 1 - sin-1 0 = Ï€/2 - 0 = Ï€/2 Hence, âˆ«01 1/âˆš(1-x2) dx = Ï€/2 -----(1) âˆ«01 1/âˆš(1-t2) dt We know that  âˆ«01 1/âˆš(1-t2) dt = sin-1 t + C â‡’ âˆ«01 1/âˆš(1-t2) dt = [sin-1 t]01 = sin-1 1 - sin-1 0 Ï€/2 - 0 = Ï€/2 Hence, âˆ«01 1/âˆš(1-t2) dt = Ï€/2 ------(2) âˆ´ âˆ«01 1/âˆš(1-x2) dx = âˆ«01 1/âˆš(1-t2) dt Property 2: âˆ«ab f(x) dx = - âˆ«ba f(x) dx Also, if b = a, then âˆ«ab f(x) dx = 0 Proof: Let âˆ« f(x) dx = G(x) Second fundamental theorem on integral calculus: If f is a continuous function defined on [a, b] and âˆ« f(x) dx = G(x), then âˆ«ab f(x) dx = G(x)ab = G(b) - G(a) âˆ«ab f(x) dx = G(x)ab     = G(b) - G(a)     = - [G(a) - G(b)]     = - [G(x)]ab     = - âˆ«ba f(x) dx âˆ«ab f(x) dx = - âˆ«ba f(x) dx If b = a, then âˆ«ba f(x) dx = âˆ«aa f(x) dx âˆ« f(x) dx = G(x) + C âˆ«aa f(x) dx = G(x)aa = G(a) - G(a) = 0 Example: I = âˆ«0Ï€  sin x/(1+cos2x) dx cos x = t â‡’ - sin x dx = dt â‡’ sin x dx = - dt When x = 0,â‡’ t = cos 0 = 1 When x = Ï€ , â‡’ t = cos Ï€ = -1 âˆ«0Ï€  sin x/(1+cos2x) dx = âˆ«1-1  -dt/(1+t2) By property on integral calculus, we have âˆ«ab f(x) dx = - âˆ«ba f(x) dx âˆ«1-1  -dt/(1+t2) = - âˆ«-11  -dt/(1+t2) âˆ«-11  dt/(1+t2) = (tan-1 t)-11             ('.'  âˆ«  dt/(1+t2) = tan-1 t) = (tan-1 1 - tan-1 (-1)) = tan-1 1 + tan-1 1 = Ï€/4 + Ï€/4 = 2Ï€/4 = Ï€/2 Hence, âˆ«0Ï€  sin x/(1+cos2x) dx = Ï€/2 Property 3: If a < c < b  then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx Proof: Let G(x) = âˆ« f(x) dx âˆ«ab f(x) dx = G(b) - G(a) â€¦.(i) âˆ«ac f(x) dx = G(c) - G(a)â€¦..(ii) âˆ«cb f(x) dx = G(b) - G(c) â€¦.(iii) (ii) + (iii) = âˆ«ac f(x) dx + âˆ«cb f(x) dx = G(c) - G(a) + G(b) - G(c) â‡’ âˆ«ac f(x) dx + âˆ«ac f(x) dx = G(b) - G(a) = âˆ«ab f(x) dx âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx Ex: Evaluate âˆ«04 |2 - x| dx Let I = âˆ«04 |2 - x| dx We know that If a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx So, âˆ«04 |2 - x| dx = âˆ«02 |2 - x| dx + âˆ«24 |2 - x| dx = âˆ«02 |2 - x| dx + âˆ«24 |2 - x| dx = âˆ«02 2 dx - âˆ«02 x dx +  âˆ«24 x dx - âˆ«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, âˆ«04 |2 - x| dx = 4

#### Summary

Properties of definite integrals: 1. âˆ«ab f(x) dx = âˆ«ab f(t) dt 2. âˆ«abf(x) dx = - âˆ«abf(x) dx 3. if a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cbf(x) dx Property 1: âˆ«ab f(x) dx = âˆ«ab f(t) dt Ex: Evaluate âˆ«01 1/âˆš(1-x2) dx Claim:  âˆ«01 1/âˆš(1-x2) dx = âˆ«01 1/âˆš(1-t2) dt We know that  âˆ« 1/âˆš(1-x2) dx = sin-1 x + C â‡’âˆ«01 1/âˆš(1-x2) dx = [sin-1 x]01 = sin-1 1 - sin-1 0 = Ï€/2 - 0 = Ï€/2 Hence, âˆ«01 1/âˆš(1-x2) dx = Ï€/2 -----(1) âˆ«01 1/âˆš(1-t2) dt We know that  âˆ«01 1/âˆš(1-t2) dt = sin-1 t + C â‡’ âˆ«01 1/âˆš(1-t2) dt = [sin-1 t]01 = sin-1 1 - sin-1 0 Ï€/2 - 0 = Ï€/2 Hence, âˆ«01 1/âˆš(1-t2) dt = Ï€/2 ------(2) âˆ´ âˆ«01 1/âˆš(1-x2) dx = âˆ«01 1/âˆš(1-t2) dt Property 2: âˆ«ab f(x) dx = - âˆ«ba f(x) dx Also, if b = a, then âˆ«ab f(x) dx = 0 Proof: Let âˆ« f(x) dx = G(x) Second fundamental theorem on integral calculus: If f is a continuous function defined on [a, b] and âˆ« f(x) dx = G(x), then âˆ«ab f(x) dx = G(x)ab = G(b) - G(a) âˆ«ab f(x) dx = G(x)ab     = G(b) - G(a)     = - [G(a) - G(b)]     = - [G(x)]ab     = - âˆ«ba f(x) dx âˆ«ab f(x) dx = - âˆ«ba f(x) dx If b = a, then âˆ«ba f(x) dx = âˆ«aa f(x) dx âˆ« f(x) dx = G(x) + C âˆ«aa f(x) dx = G(x)aa = G(a) - G(a) = 0 Example: I = âˆ«0Ï€  sin x/(1+cos2x) dx cos x = t â‡’ - sin x dx = dt â‡’ sin x dx = - dt When x = 0,â‡’ t = cos 0 = 1 When x = Ï€ , â‡’ t = cos Ï€ = -1 âˆ«0Ï€  sin x/(1+cos2x) dx = âˆ«1-1  -dt/(1+t2) By property on integral calculus, we have âˆ«ab f(x) dx = - âˆ«ba f(x) dx âˆ«1-1  -dt/(1+t2) = - âˆ«-11  -dt/(1+t2) âˆ«-11  dt/(1+t2) = (tan-1 t)-11             ('.'  âˆ«  dt/(1+t2) = tan-1 t) = (tan-1 1 - tan-1 (-1)) = tan-1 1 + tan-1 1 = Ï€/4 + Ï€/4 = 2Ï€/4 = Ï€/2 Hence, âˆ«0Ï€  sin x/(1+cos2x) dx = Ï€/2 Property 3: If a < c < b  then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx Proof: Let G(x) = âˆ« f(x) dx âˆ«ab f(x) dx = G(b) - G(a) â€¦.(i) âˆ«ac f(x) dx = G(c) - G(a)â€¦..(ii) âˆ«cb f(x) dx = G(b) - G(c) â€¦.(iii) (ii) + (iii) = âˆ«ac f(x) dx + âˆ«cb f(x) dx = G(c) - G(a) + G(b) - G(c) â‡’ âˆ«ac f(x) dx + âˆ«ac f(x) dx = G(b) - G(a) = âˆ«ab f(x) dx âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx Ex: Evaluate âˆ«04 |2 - x| dx Let I = âˆ«04 |2 - x| dx We know that If a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx So, âˆ«04 |2 - x| dx = âˆ«02 |2 - x| dx + âˆ«24 |2 - x| dx = âˆ«02 |2 - x| dx + âˆ«24 |2 - x| dx = âˆ«02 2 dx - âˆ«02 x dx +  âˆ«24 x dx - âˆ«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, âˆ«04 |2 - x| dx = 4

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