Chapter 7 Class 12 Integrals
Serial order wise

Odd function  f(x) = -f(x) 

Even function  g(x) = g(x) 

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


Question 1

1 (-1)  x 99    dx=______________.

(a) 0   (b) 1  (c) −1  (d) 2


Question 2

π x cos x dx =______________.

(a) 1   (b) 0      (c) −1      (d) π/2


Question 3

π/2 -π/2   sin 3 π‘₯ 𝑑π‘₯ = _________.

(a) 1    (b) 0     (c) −1     (d) π



Question 4

π   π‘₯ sin π‘₯ 𝑑π‘₯ = _________.

(a) π      (b) 0           (c) 2π            (d) π/2

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Question 5

π   tanπ‘₯ sec 2 π‘₯ 𝑑π‘₯ = _________.

(a) 1            (b) −1          (c) 0            (d) 2


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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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