Notes On Properties of Definite Integrals - III - CBSE Class 12 Maths
Property 6: âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx Proof: âˆ«02a f(x) dx By a property on definite integrals, if a < c < b , then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«a2a f(x) dx -----(1) Let's evaluate âˆ«02a f(x) dx Let t = 2a - x â‡’ x = 2a - t dx = 0 - dt â‡’ dx = -dt When x = a, t = 2a - a = a When x = 2a, t = 2a - 2a = 0 âˆ«02a f(x) dx = âˆ«a0 f(2a - t) x (-dt)  = - âˆ«a0 f(2a - t) dt By a property of definite integrals, âˆ«ab f(x) dx = - âˆ«ba f(x) dx    = âˆ«0a f(2a - t) dt By a property of definite integrals, âˆ«ab f(x) dx = âˆ«ab f(t) dt       =  âˆ«0a f(2a - x) dx âˆ´ âˆ«02a f(x) dx = âˆ«0a f(2a - x) dx ----(2) On substituting equation 2, in equation 1, we get âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx Example: Evaluate âˆ«0Ï€  (x sin x)/(1 + cos2x) dx. Let I = âˆ«0Ï€  (x sin x)/(1 + cos2x) dx = âˆ«02Ï€/2  (x sin x)/(1 + cos2x) dx âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx = âˆ«02Ã—Ï€/2  (x sin x)/(1 + cos2x) dx = = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin (Ï€-x))/(1 + cos2(Ï€-x)) dx  = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin x)/(1 + (- cos x )2) dx  = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  (Ï€ sin x - x sin x)/(1 + cos2 x) dx = âˆ«0Ï€/2  (x sin x + Ï€ sin x - x sin x)/(1 + cos2x) dx = âˆ«0Ï€/2  (Ï€ sin x)/(1 + cos2x) dx = Ï€ âˆ«0Ï€/2  sin x/(1 + cos2x) dx -----(1) Let cos x = t â‡’ - sin x dx = dt â‡’ sin x dx = - dt When x = Ï€/2 , t = cos Ï€/2 = 0 When x = 0 , t = cos 0 = 1 = Ï€ âˆ«10 -1/(1+t2) dt = -Ï€ âˆ«10 dt/(1+t2) âˆ«ab f(x) dx = - âˆ«ba f(x) dx  -Ï€ âˆ«10 dt/(1+t2) = = Ï€ âˆ«01 dt/(1+t2) âˆ« 1/(1+t2) dt = tan-1t + C Ï€ âˆ« dt/(1+t2) = Ï€ [tan-1t]01                   = Ï€ [tan-11 - tan-10]                   = Ï€(Ï€/4 - 0)                    = Ï€2/4 Property 7: $\underset{\text{0}}{\overset{\text{2a}}{âˆ«}}\text{f(x)}=\left\{\begin{array}{ll}2\underset{\text{0}}{\overset{\text{a}}{âˆ«}}\text{f(x)dx,}& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a -x\right) = f\left(x\right)}\\ 0,& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a - x\right)}=-f\left(x\right)\end{array}\right\$ Proof: âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx----(1) If f(2a - x) = f(x),   then from (1) , we have âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx                 = 2 . âˆ«0a f(x) dx If f(2a - x) = - f(x), then from (1), we have âˆ«02a f(x) dx = âˆ«0a f(x) dx - âˆ«0a f(x) dx                 = 0 $\underset{\text{0}}{\overset{\text{2a}}{âˆ«}}\text{f(x)}=\left\{\begin{array}{ll}2\underset{\text{0}}{\overset{\text{a}}{âˆ«}}\text{f(x)dx,}& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a -x\right) = f\left(x\right)}\\ 0,& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a - x\right)}=-f\left(x\right)\end{array}\right\$ Example: Evaluate âˆ«0Ï€ tan x/(sec x + cos x) dx Let f(x) = tan x/(sec x + cos x)  and 2a = Ï€ Let's find out f(2a - x) = f(Ï€-x) = tan (Ï€-x)/(sec (Ï€-x) + cos (Ï€-x))                                = - tanx/(-secx-cosx)                                = tanx/(secx + cosx)                                = f(x) By a property on definite integrals, âˆ«02a f(x) dx = 2. âˆ«0a f(x) dx if f(2a - x) = f(x) âˆ´ âˆ«0Ï€ tanx/(secx + cosx) dx = 2 âˆ«0Ï€/2 tanx/(secx + cosx) dx                                         = 2 âˆ«0Ï€/2 (sin x/cox x)/(1/cos x + cosx) dx                                          = 2 âˆ«0Ï€/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 = Ï€/2; t = cos Ï€/2 = 0 Therefore, 2 âˆ«0Ï€/2 sin x/(1 + cos2x) dx = 2.âˆ«10 -dt/(1+t2) âˆ«ab f(x) dx = - âˆ«ba f(x) dx 2 .âˆ«10 -dt/(1+t2) = 2 .âˆ«01 1/(1+t2) dt                        = 2 [tan-1 t]01                         = 2[tan-1(1) - tan-1(0)]                         = 2[Ï€/4 - 0]                         = Ï€/2 âˆ«0Ï€ tanx/(sec x + cos x) dx = = Ï€/2 Property 8: i. âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx, if f is an even function , i.e, f(-x) = f(x) ii. âˆ«-aa f(x) dx = 0 if f is an odd function , i.e, f(-x) = - f(x) Proof: i. LHS = âˆ«-aa f(x) dx If a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx We have - a < 0 < a and âˆ«-aa f(x) dx = âˆ«-a0 f(x) dx + âˆ«0a f(x) dx  ------(1) âˆ«-a0 f(x) dx Let x = -t dx = -dt When x = -a, then -t = a â‡’ t = a When x = 0 , then -t = 0 â‡’ t = 0 âˆ«-a0 f(x) dx = âˆ«-a0 f(-t) (-dt)  = - âˆ«a0 f(-t) dt  = âˆ«0a f(-t) dt      ['.' âˆ«ab f(x) dx = - âˆ«ba f(x) dx] âˆ«-a0 f(x) dx = âˆ«0a f(-t) dt ----(2) However, by a property of definite integrals, âˆ«ab f(x) dx = âˆ«ab f(t) dt âˆ«-a0 f(x) dx = âˆ«0a f(-x) dx -----(3)   âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx -----(4) Case - 1:If the function is an even function, i.e. f(-x) = f(x): âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx Case - 2: If the function is an odd function, i.e. f(-x) = - f(x): â‡’ âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = âˆ«0a -f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = - âˆ«0a f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = 0 i. âˆ«-aa f(x) dx = 2. âˆ«0a f(x) dx, if f is an even function , i.e f(-x) = f(x) ii. âˆ«-aa f(x) dx =  0 if f is an odd function , i.e f(-x) = -f(x) Example: Evaluate âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx = âˆ«-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) â‡’ f(-x) = - f(x) âˆ´ f(x) is an odd function âˆ«-aa f(x) dx =  0 if f is an odd function , i.e, f(-x) = - f(x) âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx = 0

#### Summary

Property 6: âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx Proof: âˆ«02a f(x) dx By a property on definite integrals, if a < c < b , then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«a2a f(x) dx -----(1) Let's evaluate âˆ«02a f(x) dx Let t = 2a - x â‡’ x = 2a - t dx = 0 - dt â‡’ dx = -dt When x = a, t = 2a - a = a When x = 2a, t = 2a - 2a = 0 âˆ«02a f(x) dx = âˆ«a0 f(2a - t) x (-dt)  = - âˆ«a0 f(2a - t) dt By a property of definite integrals, âˆ«ab f(x) dx = - âˆ«ba f(x) dx    = âˆ«0a f(2a - t) dt By a property of definite integrals, âˆ«ab f(x) dx = âˆ«ab f(t) dt       =  âˆ«0a f(2a - x) dx âˆ´ âˆ«02a f(x) dx = âˆ«0a f(2a - x) dx ----(2) On substituting equation 2, in equation 1, we get âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx Example: Evaluate âˆ«0Ï€  (x sin x)/(1 + cos2x) dx. Let I = âˆ«0Ï€  (x sin x)/(1 + cos2x) dx = âˆ«02Ï€/2  (x sin x)/(1 + cos2x) dx âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx = âˆ«02Ã—Ï€/2  (x sin x)/(1 + cos2x) dx = = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin (Ï€-x))/(1 + cos2(Ï€-x)) dx  = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  ((Ï€- x) sin x)/(1 + (- cos x )2) dx  = âˆ«0Ï€/2  (x sin x)/(1 + cos2x) dx +  âˆ«0Ï€/2  (Ï€ sin x - x sin x)/(1 + cos2 x) dx = âˆ«0Ï€/2  (x sin x + Ï€ sin x - x sin x)/(1 + cos2x) dx = âˆ«0Ï€/2  (Ï€ sin x)/(1 + cos2x) dx = Ï€ âˆ«0Ï€/2  sin x/(1 + cos2x) dx -----(1) Let cos x = t â‡’ - sin x dx = dt â‡’ sin x dx = - dt When x = Ï€/2 , t = cos Ï€/2 = 0 When x = 0 , t = cos 0 = 1 = Ï€ âˆ«10 -1/(1+t2) dt = -Ï€ âˆ«10 dt/(1+t2) âˆ«ab f(x) dx = - âˆ«ba f(x) dx  -Ï€ âˆ«10 dt/(1+t2) = = Ï€ âˆ«01 dt/(1+t2) âˆ« 1/(1+t2) dt = tan-1t + C Ï€ âˆ« dt/(1+t2) = Ï€ [tan-1t]01                   = Ï€ [tan-11 - tan-10]                   = Ï€(Ï€/4 - 0)                    = Ï€2/4 Property 7: $\underset{\text{0}}{\overset{\text{2a}}{âˆ«}}\text{f(x)}=\left\{\begin{array}{ll}2\underset{\text{0}}{\overset{\text{a}}{âˆ«}}\text{f(x)dx,}& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a -x\right) = f\left(x\right)}\\ 0,& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a - x\right)}=-f\left(x\right)\end{array}\right\$ Proof: âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a - x) dx----(1) If f(2a - x) = f(x),   then from (1) , we have âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx                 = 2 . âˆ«0a f(x) dx If f(2a - x) = - f(x), then from (1), we have âˆ«02a f(x) dx = âˆ«0a f(x) dx - âˆ«0a f(x) dx                 = 0 $\underset{\text{0}}{\overset{\text{2a}}{âˆ«}}\text{f(x)}=\left\{\begin{array}{ll}2\underset{\text{0}}{\overset{\text{a}}{âˆ«}}\text{f(x)dx,}& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a -x\right) = f\left(x\right)}\\ 0,& \hfill \phantom{\rule{10}{0ex}}\mathrm{if f\left(2a - x\right)}=-f\left(x\right)\end{array}\right\$ Example: Evaluate âˆ«0Ï€ tan x/(sec x + cos x) dx Let f(x) = tan x/(sec x + cos x)  and 2a = Ï€ Let's find out f(2a - x) = f(Ï€-x) = tan (Ï€-x)/(sec (Ï€-x) + cos (Ï€-x))                                = - tanx/(-secx-cosx)                                = tanx/(secx + cosx)                                = f(x) By a property on definite integrals, âˆ«02a f(x) dx = 2. âˆ«0a f(x) dx if f(2a - x) = f(x) âˆ´ âˆ«0Ï€ tanx/(secx + cosx) dx = 2 âˆ«0Ï€/2 tanx/(secx + cosx) dx                                         = 2 âˆ«0Ï€/2 (sin x/cox x)/(1/cos x + cosx) dx                                          = 2 âˆ«0Ï€/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 = Ï€/2; t = cos Ï€/2 = 0 Therefore, 2 âˆ«0Ï€/2 sin x/(1 + cos2x) dx = 2.âˆ«10 -dt/(1+t2) âˆ«ab f(x) dx = - âˆ«ba f(x) dx 2 .âˆ«10 -dt/(1+t2) = 2 .âˆ«01 1/(1+t2) dt                        = 2 [tan-1 t]01                         = 2[tan-1(1) - tan-1(0)]                         = 2[Ï€/4 - 0]                         = Ï€/2 âˆ«0Ï€ tanx/(sec x + cos x) dx = = Ï€/2 Property 8: i. âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx, if f is an even function , i.e, f(-x) = f(x) ii. âˆ«-aa f(x) dx = 0 if f is an odd function , i.e, f(-x) = - f(x) Proof: i. LHS = âˆ«-aa f(x) dx If a < c < b, then âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx We have - a < 0 < a and âˆ«-aa f(x) dx = âˆ«-a0 f(x) dx + âˆ«0a f(x) dx  ------(1) âˆ«-a0 f(x) dx Let x = -t dx = -dt When x = -a, then -t = a â‡’ t = a When x = 0 , then -t = 0 â‡’ t = 0 âˆ«-a0 f(x) dx = âˆ«-a0 f(-t) (-dt)  = - âˆ«a0 f(-t) dt  = âˆ«0a f(-t) dt      ['.' âˆ«ab f(x) dx = - âˆ«ba f(x) dx] âˆ«-a0 f(x) dx = âˆ«0a f(-t) dt ----(2) However, by a property of definite integrals, âˆ«ab f(x) dx = âˆ«ab f(t) dt âˆ«-a0 f(x) dx = âˆ«0a f(-x) dx -----(3)   âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx -----(4) Case - 1:If the function is an even function, i.e. f(-x) = f(x): âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx Case - 2: If the function is an odd function, i.e. f(-x) = - f(x): â‡’ âˆ«-aa f(x) dx = âˆ«0a f(-x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = âˆ«0a -f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = - âˆ«0a f(x) dx + âˆ«0a f(x) dx â‡’ âˆ«-aa f(x) dx = 0 i. âˆ«-aa f(x) dx = 2. âˆ«0a f(x) dx, if f is an even function , i.e f(-x) = f(x) ii. âˆ«-aa f(x) dx =  0 if f is an odd function , i.e f(-x) = -f(x) Example: Evaluate âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx = âˆ«-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) â‡’ f(-x) = - f(x) âˆ´ f(x) is an odd function âˆ«-aa f(x) dx =  0 if f is an odd function , i.e, f(-x) = - f(x) âˆ«logÂ½log2 sin ((ex-1)/(ex+1))dx = 0

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