Evaluate
∫ (x 3 + 1) dx from -2 to 2
![Question 17 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2](https://cdn.teachoo.com/ceb5d3ea-875a-47e2-b6f8-a135432cd781/slide43.jpg)
![Question 17 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3](https://cdn.teachoo.com/78b3ef80-3ad7-4a4e-8d39-8e06099a6721/slide44.jpg)
CBSE Class 12 Sample Paper for 2020 Boards
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at Dec. 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