Evaluate
∫ (x ^{ 3 } + 1) dx from -2 to 2
CBSE Class 12 Sample Paper for 2020 Boards
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13
Question 14 (OR 1st Question)
Question 14 (OR 2nd Question) Important
Question 15 (OR 1st Question)
Question 15 (OR 2nd Question) Important
Question 16
Question 17 Important You are here
Question 18 (OR 1st Question)
Question 18 (OR 2nd Question)
Question 19
Question 20
Question 21 (OR 1st Question) Important
Question 21 (OR 2nd Question) Important
Question 22
Question 23
Question 24 (OR 1st Question)
Question 24 (OR 2nd Question)
Question 25
Question 26 Important
Question 27
Question 28 (OR 1st Question) Important
Question 28 (OR 2nd Question)
Question 29
Question 30 Important
Question 31 (OR 1st Question) Important
Question 31 (OR 2nd Question)
Question 32 Important
Question 33 (OR 1st Question) Important
Question 33 (OR 2nd Question) Important
Question 34
Question 35 (OR 1st Question) Important
Question 35 (OR 2nd Question) Important
Question 36 Important
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at April 16, 2024 by Teachoo
Question 17 ∫ (x3 + 1) dx from -2 to 2 This is of form ∫ a -a f (x) dx And we now that ∫_(−𝑎)𝑎 𝑓(𝑥)𝑑𝑥=0,〗 if f(−𝑥)=−𝑓(𝑥) And ∫_(−𝑎)^𝑎𝑓(𝑥)𝑑𝑥=2∫_0^𝑎𝑓(𝑥)𝑑𝑥 , if f(−𝑥)=𝑓(𝑥) Now, ∫_(−2)^2(𝑥^3+1)𝑑𝑥 = ∫_(−2)^2 𝑥^3 𝑑𝑥〗 + ∫_(−2)^21𝑑𝑥 Since 〖(−𝑥)〗^3=−𝑥^3 So, ∫_(−𝑎)^𝑎▒〖𝑥^3 𝑑𝑥=0,〗 And 1 is constant, so f(–x) = f(x) ∫_(−𝑎)^𝑎▒1𝑑𝑥=2∫_0^𝑎▒𝑑𝑥 = 0 + 2∫_0^2▒1𝑑𝑥 = 2∫_0^2▒𝑑𝑥 = 2 〖[𝑥]〗_0^2 = 2 (2 – 0) = 4