Question 11 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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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