Check sibling questions
Chapter 7 Class 12 Integrals
Serial order wise

Odd functionΒ  f(x) = -f(x)Β 

Even functionΒ  g(x) = g(x)Β 

Based on the above information, answer any fourΒ  of the following questions.

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Question 1

∫ 1 (-1)  x 99    dx=______________.

(a) 0 Β  (b) 1Β  (c) βˆ’1Β  (d) 2

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Question 2

∫ Ο€ -Ο€ x cos x dx =______________.

(a) 1 Β  (b) 0Β  Β  Β  (c) βˆ’1Β  Β  Β  (d) Ο€/2

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Question 3

∫ Ο€/2 -Ο€/2 Β  sin 3 π‘₯ 𝑑π‘₯ = _________.

(a) 1Β  Β  (b) 0Β  Β  Β (c) βˆ’1Β  Β  Β (d) Ο€

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Β 

Question 4

∫ Ο€ -Ο€ Β  π‘₯ sin π‘₯ 𝑑π‘₯ = _________.

(a) π  Β  Β  (b) 0Β  Β  Β  Β  Β  Β (c) 2π  Β  Β  Β  Β  Β  (d) Ο€/2

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Question 5

∫ Ο€ -Ο€ Β  tanπ‘₯ sec 2 π‘₯ 𝑑π‘₯ = _________.

(a) 1Β  Β  Β  Β  Β  Β  (b) βˆ’1Β  Β  Β  Β  Β  (c) 0Β  Β  Β  Β  Β  Β  (d) 2

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Transcript

Question 1 Based on the above information, answer any four of the following questions. Question 1 ∫1_(βˆ’1)^1β–’π‘₯^99 𝑑π‘₯=______________. (a) 0 (b) 1 (c) βˆ’1 (d) 2 This is of form ∫_(βˆ’π’‚)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=π‘₯^99 𝒇(βˆ’π’™)=(βˆ’π‘₯)^99=βˆ’π‘₯^99 Thus, 𝒇(βˆ’π’™) =βˆ’π’‡β‘(𝒙) ∫1_(βˆ’1)^1β–’π‘₯^99 𝑑𝒙= 0 So, the correct answer is (a) Question 2 ∫1_(βˆ’πœ‹)^πœ‹β–’γ€– π‘₯ cos⁑π‘₯ γ€— 𝑑π‘₯=______________. (a) 1 (b) 0 (c) βˆ’1 (d) πœ‹/2 This is of form ∫_(βˆ’π’‚)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=π‘₯ π‘π‘œπ‘  π‘₯ 𝒇(βˆ’π’™)=(βˆ’π‘₯) cos⁑〖(βˆ’π‘₯)γ€—=βˆ’π‘₯ cos⁑π‘₯ Thus, 𝒇(βˆ’π’™) =βˆ’π’‡β‘(𝒙) ∫1_(βˆ’πœ‹)^πœ‹β–’γ€– π‘₯ cos⁑π‘₯ γ€— 𝑑π‘₯= 0 So, the correct answer is (b) Question 3 ∫1_((βˆ’πœ‹)/2)^(πœ‹/2 )β–’γ€– sinγ€—^3⁑π‘₯ 𝑑π‘₯ = _________. (a) 1 (b) 0 (c) βˆ’1 (d) πœ‹ This is of form ∫_(βˆ’π’‚)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=sin^3⁑π‘₯ 𝒇(βˆ’π’™)=sin^3⁑〖(βˆ’π‘₯)γ€—=(βˆ’sin⁑π‘₯ )^3=βˆ’sin^3⁑π‘₯ Thus, 𝒇(βˆ’π’™) =βˆ’π’‡β‘(𝒙) ∫1_((βˆ’πœ‹)/2)^(πœ‹/2 )β–’γ€– sinγ€—^3⁑π‘₯ 𝑑π‘₯ = 0 So, the correct answer is (b) Question 4 ∫1_(βˆ’πœ‹)^πœ‹β–’γ€–π‘₯ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ = _________. (a) πœ‹ (b) 0 (c) 2πœ‹ (d) πœ‹/2 This is of form ∫_(βˆ’π’‚)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=π‘₯ 𝑠𝑖𝑛 π‘₯ 𝒇(βˆ’π’™)=(βˆ’π‘₯) 〖𝑠𝑖𝑛 〗⁑〖(βˆ’π‘₯)γ€—=βˆ’π‘₯ Γ— βˆ’sin⁑π‘₯=π‘₯ sin⁑〖π‘₯ γ€— Thus, 𝒇(βˆ’π’™) =𝒇⁑(𝒙) Therefore, ∫1_(βˆ’πœ‹)^πœ‹β–’γ€–π‘₯ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ = 2∫1_0^πœ‹β–’γ€–π‘₯ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ Let I = 𝟐∫1_𝟎^𝝅▒〖𝒙 π’”π’Šπ’ 〗⁑𝒙 𝑑π‘₯ I = 2∫1_0^πœ‹β–’γ€–(𝝅 βˆ’π’™) 𝑠𝑖𝑛 〗⁑〖(πœ‹ βˆ’π‘₯) γ€— 𝑑π‘₯ I = 2∫1_0^πœ‹β–’γ€–(πœ‹ βˆ’π‘₯) 𝑠𝑖𝑛 〗⁑〖π‘₯ γ€— 𝑑π‘₯ I = 𝟐∫1_𝟎^π…β–’γ€–πœ‹ π’”π’Šπ’ 〗⁑𝒙 𝑑π‘₯ βˆ’ 𝟐∫1_𝟎^𝝅▒〖𝒙 π’”π’Šπ’ 〗⁑𝒙 𝑑π‘₯ Adding (1) and (2) I + I = 2∫1_0^πœ‹β–’γ€–π‘₯ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ + 2∫1_0^πœ‹β–’γ€–πœ‹ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ βˆ’ 2∫1_0^πœ‹β–’γ€–π‘₯ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ 2I = 2∫1_0^πœ‹β–’γ€–πœ‹ 𝑠𝑖𝑛 〗⁑π‘₯ 𝑑π‘₯ I = ∫1_𝟎^𝝅▒〖𝝅 π’”π’Šπ’ 〗⁑𝒙 𝑑π‘₯ I = πœ‹βˆ«1_0^πœ‹β–’γ€–π‘ π‘–π‘› 〗⁑π‘₯ 𝑑π‘₯ I = 𝝅〖[βˆ’πœπ¨π¬β‘π’™]γ€—_𝟎^𝝅 I = πœ‹[βˆ’cosβ‘γ€–πœ‹ βˆ’(βˆ’cos⁑0)γ€—] I = πœ‹[βˆ’cosβ‘πœ‹+cos⁑0] I = πœ‹[βˆ’(βˆ’1)+1] I = πœ‹ [1+1] I = πŸπ… So, the correct answer is (c) Question 5 ∫1_(βˆ’πœ‹)^πœ‹β–’γ€–tan⁑π‘₯ sec^2⁑π‘₯ γ€— 𝑑π‘₯ = _________. (a) 1 (b) βˆ’1 (c) 0 (d) 2 This is of form ∫_(βˆ’π’‚)^𝒂▒𝒇(𝒙)𝒅𝒙 𝒇(𝒙)=tan⁑π‘₯ sec^2⁑π‘₯ 𝒇(βˆ’π’™)=tan⁑〖(βˆ’π‘₯)γ€— sec^2⁑〖(βˆ’π‘₯)γ€—=βˆ’tan⁑π‘₯ sec^2⁑〖π‘₯ γ€— Thus, 𝒇(βˆ’π’™) =βˆ’π’‡β‘(𝒙) ∫1_(βˆ’πœ‹)^πœ‹β–’γ€–tan⁑π‘₯ sec^2⁑π‘₯ γ€— " 𝑑π‘₯"= 0 So, the correct answer is (c)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.