Odd function f(x) = -f(x) Even fu

calculus integral trick with even \u0026 odd function
calculus integral trick with even \u0026 odd function

Question 1 – Case Based Questions (MCQ) – Chapter 7 Class 12 Integrals

Last updated at May 29, 2023 by Teachoo

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

𝑥 𝑑𝑥 = _________.

(a) 1 (b) 0 (c) −1 (d) π


Question 4



π




𝑥 sin 𝑥 𝑑𝑥 = _________.

(a) π (b) 0 (c) 2π (d) π/2


Question 5



π




tan𝑥 sec

2

𝑥 𝑑𝑥 = _________.

(a) 1 (b) −1 (c) 0 (d) 2

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Transcript

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 cos𝑥 〗 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 cos𝑥
Thus, 𝒇(−𝒙) =−𝒇(𝒙)
∫1_(−𝜋)^𝜋▒〖 𝑥 cos𝑥 〗 𝑑𝑥= 0
So, the correct answer is (b)
Question 3 ∫1_((−𝜋)/2)^(𝜋/2 )▒〖 sin〗^3𝑥 𝑑𝑥 = _________. (a) 1 (b) 0 (c) −1 (d) 𝜋
This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙
𝒇(𝒙)=sin^3𝑥
𝒇(−𝒙)=sin^3〖(−𝑥)〗=(−sin𝑥 )^3=−sin^3𝑥
Thus, 𝒇(−𝒙) =−𝒇(𝒙)
∫1_((−𝜋)/2)^(𝜋/2 )▒〖 sin〗^3𝑥 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 × −sin𝑥=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〖[−𝐜𝐨𝐬𝒙]〗_𝟎^𝝅
I = 𝜋[−cos〖𝜋 −(−cos0)〗]
I = 𝜋[−cos𝜋+cos0]
I = 𝜋[−(−1)+1]
I = 𝜋 [1+1]
I = 𝟐𝝅
So, the correct answer is (c)
Question 5 ∫1_(−𝜋)^𝜋▒〖tan𝑥 sec^2𝑥 〗 𝑑𝑥 = _________. (a) 1 (b) −1 (c) 0 (d) 2
This is of form ∫_(−𝒂)^𝒂▒𝒇(𝒙)𝒅𝒙
𝒇(𝒙)=tan𝑥 sec^2𝑥
𝒇(−𝒙)=tan〖(−𝑥)〗 sec^2〖(−𝑥)〗=−tan𝑥 sec^2〖𝑥 〗
Thus, 𝒇(−𝒙) =−𝒇(𝒙)
∫1_(−𝜋)^𝜋▒〖tan𝑥 sec^2𝑥 〗 ” 𝑑𝑥”= 0
So, the correct answer is (c)

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