Evaluate ∫_(-1)^2|x^3-3x^2+2x| dx
This question is similar to Question 30 - CBSE Class 12 Sample Paper 2020 Boards



CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Last updated at Dec. 14, 2024 by Teachoo
This question is similar to Question 30 - CBSE Class 12 Sample Paper 2020 Boards
Transcript
Question 11 Evaluate ∫_(−1)^2▒|𝑥^3−3𝑥^2+2𝑥| dx |𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙|=|𝑥(𝑥^2−3𝑥+2)| =|𝑥(𝑥^2−2𝑥−𝑥+2)| =|𝑥(𝑥(𝑥−2)−1(𝑥−2)) | =|𝒙(𝒙−𝟏)(𝒙−𝟐)| Thus, 𝑥=0,𝑥=1,𝑥=2 ∴ |𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙|={█(−𝑥 . −(𝑥−1) . −(𝑥−2) 𝑖𝑓 −1≤𝑥<0@𝑥 . −(𝑥−1) . −(𝑥−2) 𝑖𝑓 0≤𝑥<1@𝑥 . (𝑥−1) . −(𝑥−2) 𝑖𝑓 1≤𝑥<2)┤ ={█(−𝑥(𝑥−1) (𝑥−2) 𝑖𝑓 −1≤𝑥<0@𝑥(𝑥−1)(𝑥−2) 𝑖𝑓 0≤𝑥<1@−𝑥(𝑥−1)(𝑥−2) 𝑖𝑓 1≤𝑥<2)┤ ={█(−(𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙) 𝑖𝑓 −1≤𝑥<0@(𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙) 𝑖𝑓 0≤𝑥<1@−(𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙) 𝑖𝑓 1≤𝑥<2)┤ Now, ∫_(−𝟏)^𝟐▒|𝒙^𝟑−𝟑𝒙^𝟐+𝟐𝒙| dx = ∫_(−1)^0▒〖−(𝑥^3−3𝑥^2+2𝑥)〗 𝑑𝑥+∫_0^1▒〖(𝑥^3−3𝑥^2+2𝑥)〗 𝑑𝑥 +∫_1^2▒〖−(𝑥^3−3𝑥^2+2𝑥)〗 𝑑𝑥 = −[𝑥^4/4−3 ×𝑥^3/3+2 ×𝑥^2/2]_(−1)^0+[𝑥^4/4−3 ×𝑥^3/3+2 ×𝑥^2/2]_0^1 ` −[𝑥^4/4−3 ×𝑥^3/3+2 ×𝑥^2/2]_1^2 = −[𝒙^𝟒/𝟒−𝒙^𝟑+𝒙^𝟐 ]_(−𝟏)^𝟎+[𝒙^𝟒/𝟒−𝒙^𝟑+𝒙^𝟐 ]_𝟎^𝟏−[𝒙^𝟒/𝟒−𝒙^𝟑+𝒙^𝟐 ]_𝟏^𝟐 = −[((0^4 )/4−0^3+0^2 )−((−1)^4/4−(−1)^3+(−1)^2 )] +[(1^4/4−1^3+1^2 )−((0^4 )/4−0^3+0^2 )] −[(2^4/4−2^3+2^2 )−(1^4/4−1^3+1^2 )]= −[0−(1/4+1+1)] +[(1/4−1+1)−0] −[(4−8+4)−(1/4−1+1)] = −[−9/4]+[1/4]−[0−1/4] = 9/4+1/4+1/4 = 𝟏𝟏/𝟒