Question 27

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Question 27 Evaluate the following ∫1_((−𝜋)/4)^(𝜋/4)▒(𝑥+𝜋/4)/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_((−𝜋)/4)^(𝜋/4)▒(𝑥+𝜋/4)/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_((−𝜋)/4)^(𝜋/4)▒𝑥/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_((−𝜋)/4)^(𝜋/4)▒(𝜋/4)/(2 −cos⁡2𝑥 ) 𝑑𝑥 Let f(x) = 𝑥/(2 − cos⁡2𝑥 ) f(–x) = (−𝑥)/(2 − cos⁡〖(−2𝑥)〗 ) As cos (–x) = cos x f(–x) = (−𝑥)/(2 − cos⁡𝑥 ) ∴ f(–x) = – f(x) Let f(x) = (𝜋/4)/(2 − cos⁡2𝑥 ) f(–x) = (𝜋/4)/(2 − cos⁡〖(−2𝑥)〗 ) As cos (–x) = cos x f(–x) = (𝜋/4)/(2 − cos⁡𝑥 ) ∴ f(–x) = f(x) Using property If f(-x) = – f(x) ∫_(−𝑎)^𝑎▒〖𝑓(𝑥)𝑑𝑥〗=0 Using property If f(-x) = f(x) ∫_(−𝑎)^𝑎▒〖𝑓(𝑥)𝑑𝑥〗=2∫_0^𝑎▒𝑓(𝑥)𝑑𝑥 ∫1_((−𝜋)/4)^(𝜋/4)▒𝑥/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_((−𝜋)/4)^(𝜋/4)▒(𝜋/4)/(2 −cos⁡2𝑥 ) 𝑑𝑥 Thus, I = I1 + I2 I ∫1_0^(𝜋/4)▒(𝜋/4)/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_0^(𝜋/4)▒1/(2 −cos⁡2𝑥 ) 𝑑𝑥 ∫1_0^(𝜋/4)▒1/(2 −(1−2 sin^2⁡𝑥) ) 𝑑𝑥 ∫1_0^(𝜋/4)▒1/(1+2 sin^2⁡𝑥 ) 𝑑𝑥 ∫1_0^(𝜋/4)▒(1/cos^2⁡𝑥 )/((1+2 sin^2⁡𝑥)/cos^2⁡𝑥 ) 𝑑𝑥 ∫1_0^(𝜋/4)▒sec^2⁡𝑥/(1+tan^2⁡𝑥+2 tan^2⁡𝑥 ) 𝑑𝑥 Let tan x = t sec2 x dx = dt As x = 0 t = tan 0 = 0 As x = 𝜋/4, t = tan 𝜋/4 = 1 ∫1_0^1▒𝑑𝑡/3(1/3+t^2 ) ∫1_0^1▒𝑑𝑡/((1/√3)^2+t^2 ) Using ∫1▒〖𝑑𝑥/(𝑎^2 + 𝑥^2 )=〖1/𝑎 tan^(−1)〗⁡〖𝑥/𝑎〗 〗 = 𝜋/6 [〖1/((1/√3) ) tan^(−1)〗⁡〖𝑡/((1/√3) )〗 ]_0^1 = (√3 𝜋)/6 [tan^(−1)⁡〖√3(1)〗−tan^(−1)⁡〖√3(0)〗 ] = (√3 𝜋)/6 [tan^(−1)⁡√3−tan^(−1)⁡0 ] = (√3 𝜋)/6 [𝜋/3−0] = (√3 𝜋)/6 [𝜋/3] = (√𝟑 𝝅^𝟐)/𝟏𝟖 Question 27 Evaluate as the limit of a sum. ∫1_(−2)^2▒〖(3𝑥^2−2𝑥+4)〗 We know that ∫1▒𝑓(𝑥) 𝑑𝑥 =(𝑏−𝑎) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(𝑎)+𝑓(𝑎+ℎ)+𝑓(𝑎+2ℎ)…+𝑓(𝑎+(𝑛−1)ℎ)) Putting 𝑎 =−2 𝑏 =2 ℎ=(2 − (−2))/𝑛 =(2 + 2)/𝑛 =4/𝑛 Now, 𝑓(𝑥)=3𝑥^2−2𝑥+4 𝑓(−2)=3(−2)^2−2(−2)+4=3(4)+4+4=20 𝑓(−2+ℎ)=3〖(−2+ℎ)〗^2−2(−2+ℎ)+4 =3(4+ℎ^2−4ℎ)+4−2ℎ+4 =3ℎ^2−14ℎ+20 𝑓 (−2+2ℎ)=3〖(−2+2ℎ)〗^2−2(−2+2ℎ)+4 =3(4+〖4ℎ〗^2−8ℎ)+4−4ℎ+4 " "=12ℎ^2−28ℎ+20 𝑓 (−2+3ℎ)=3〖(−2+3ℎ)〗^2−2(−2+3ℎ)+4 =3(4+〖9ℎ〗^2−12ℎ)+4−6ℎ+4 " "=27ℎ^2−42ℎ+20 …. …… 𝑓(−2+(𝑛−1)ℎ)=3〖(−2+(𝑛−1)ℎ)〗^2−2(−2+(𝑛−1)ℎ)+4 =3(4+〖(𝑛−1)^2 ℎ〗^2−4(𝑛−1)ℎ)+4−2(𝑛−1)ℎ+4 " "=〖3(𝑛−1)〗^2 ℎ^2−14(𝑛−1)ℎ+20 Hence we can write it as =(2−(−2)) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 20+(3ℎ^2−14ℎ+20)+(12ℎ^2−28ℎ+20)+ 20+(3ℎ^2−14ℎ+20)+(12ℎ^2−28ℎ+20)+ (27ℎ^2−42ℎ+20) + … …. +(〖3(𝑛−1)〗^2 ℎ^2−14(𝑛−1)ℎ+20) + (3ℎ^2+12ℎ^2+27ℎ^2+ ………(3(𝑛−1)^2 ℎ^2 ) – (14ℎ+28ℎ+42ℎ+ …..(𝑛−1)ℎ ) + 3h^2 (1+4+9+ ……+(𝑛−1)^2 ) – 14h(1+2+3+ ……..+(𝑛−1)) + 3h^2 (1^2+2^2+3^3+ ……+(𝑛−1)^2 ) – 14h(1+2+3+ ……..+(𝑛−1)) We know that 1^2+2^2+ …+𝑛^2= (𝑛 (𝑛 + 1)(2𝑛 + 1))/6 1^2+2^2+ ……+(𝑛−1)^2 = ((𝑛 − 1) (𝑛 −1 + 1)(2(𝑛 − 1) + 1))/6 = ((𝑛 − 1) 𝑛 (2𝑛 − 2 + 1) )/6 = (𝑛 (𝑛 − 1) (2𝑛 − 1) )/6 We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+(𝑛−1) = ((𝑛 − 1) (𝑛 −1 + 1))/2 = (𝑛 (𝑛 − 1) )/2 𝑛+3h^2 ×(𝑛(𝑛 − 1)(2𝑛 − 1))/6−14ℎ ×(𝑛(𝑛 − 1))/2 Putting h = 4/𝑛 =4 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 =4 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 =4 (𝑙𝑖𝑚)┬(𝑛→∞) =4 (𝑙𝑖𝑚)┬(𝑛→∞) =4 =4 =4 =4(20+16−28) =4 ×8 =𝟑𝟐