Question 1 - Case Based Questions (MCQ) - Chapter 7 Class 12 Integrals (Term 2)

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Question 1 Based on the above information, answer any four of the following questions. Question 1 ∫1_(−1)^1▒𝑥^99 𝑑𝑥=______________. (a) 0 (b) 1 (c) −1 (d) 2 This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=𝑥^99 𝒇(−𝒙)=(−𝑥)^99=−𝑥^99 Thus, 𝒇(−𝒙) =−𝒇⁡(𝒙) ∫1_(−1)^1▒𝑥^99 𝑑𝒙= 0 So, the correct answer is (a) Question 2 ∫1_(−𝜋)^𝜋▒〖 𝑥 cos⁡𝑥 〗 𝑑𝑥=______________. (a) 1 (b) 0 (c) −1 (d) 𝜋/2 This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=𝑥 𝑐𝑜𝑠 𝑥 𝒇(−𝒙)=(−𝑥) cos⁡〖(−𝑥)〗=−𝑥 cos⁡𝑥 Thus, 𝒇(−𝒙) =−𝒇⁡(𝒙) ∫1_(−𝜋)^𝜋▒〖 𝑥 cos⁡𝑥 〗 𝑑𝑥= 0 So, the correct answer is (b) Question 3 ∫1_((−𝜋)/2)^(𝜋/2 )▒〖 sin〗^3⁡𝑥 𝑑𝑥 = _________. (a) 1 (b) 0 (c) −1 (d) 𝜋 This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=sin^3⁡𝑥 𝒇(−𝒙)=sin^3⁡〖(−𝑥)〗=(−sin⁡𝑥 )^3=−sin^3⁡𝑥 Thus, 𝒇(−𝒙) =−𝒇⁡(𝒙) ∫1_((−𝜋)/2)^(𝜋/2 )▒〖 sin〗^3⁡𝑥 𝑑𝑥 = 0 So, the correct answer is (b) Question 4 ∫1_(−𝜋)^𝜋▒〖𝑥 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 = _________. (a) 𝜋 (b) 0 (c) 2𝜋 (d) 𝜋/2 This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=𝑥 𝑠𝑖𝑛 𝑥 𝒇(−𝒙)=(−𝑥) 〖𝑠𝑖𝑛 〗⁡〖(−𝑥)〗=−𝑥 × −sin⁡𝑥=𝑥 sin⁡〖𝑥 〗 Thus, 𝒇(−𝒙) =𝒇⁡(𝒙) Therefore, ∫1_(−𝜋)^𝜋▒〖𝑥 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 = 2∫1_0^𝜋▒〖𝑥 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 Let I = 𝟐∫1_𝟎^𝝅▒〖𝒙 𝒔𝒊𝒏 〗⁡𝒙 𝑑𝑥 I = 2∫1_0^𝜋▒〖(𝝅 −𝒙) 𝑠𝑖𝑛 〗⁡〖(𝜋 −𝑥) 〗 𝑑𝑥 I = 2∫1_0^𝜋▒〖(𝜋 −𝑥) 𝑠𝑖𝑛 〗⁡〖𝑥 〗 𝑑𝑥 I = 𝟐∫1_𝟎^𝝅▒〖𝜋 𝒔𝒊𝒏 〗⁡𝒙 𝑑𝑥 − 𝟐∫1_𝟎^𝝅▒〖𝒙 𝒔𝒊𝒏 〗⁡𝒙 𝑑𝑥 Adding (1) and (2) I + I = 2∫1_0^𝜋▒〖𝑥 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 + 2∫1_0^𝜋▒〖𝜋 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 − 2∫1_0^𝜋▒〖𝑥 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 2I = 2∫1_0^𝜋▒〖𝜋 𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 I = ∫1_𝟎^𝝅▒〖𝝅 𝒔𝒊𝒏 〗⁡𝒙 𝑑𝑥 I = 𝜋∫1_0^𝜋▒〖𝑠𝑖𝑛 〗⁡𝑥 𝑑𝑥 I = 𝝅〖[−𝐜𝐨𝐬⁡𝒙]〗_𝟎^𝝅 I = 𝜋[−cos⁡〖𝜋 −(−cos⁡0)〗] I = 𝜋[−cos⁡𝜋+cos⁡0] I = 𝜋[−(−1)+1] I = 𝜋 [1+1] I = 𝟐𝝅 So, the correct answer is (c) Question 5 ∫1_(−𝜋)^𝜋▒〖tan⁡𝑥 sec^2⁡𝑥 〗 𝑑𝑥 = _________. (a) 1 (b) −1 (c) 0 (d) 2 This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=tan⁡𝑥 sec^2⁡𝑥 𝒇(−𝒙)=tan⁡〖(−𝑥)〗 sec^2⁡〖(−𝑥)〗=−tan⁡𝑥 sec^2⁡〖𝑥 〗 Thus, 𝒇(−𝒙) =−𝒇⁡(𝒙) ∫1_(−𝜋)^𝜋▒〖tan⁡𝑥 sec^2⁡𝑥 〗 " 𝑑𝑥"= 0 So, the correct answer is (c)