Example 28 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Examples
Example 1 (i)
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 28 Evaluate ∫_(−1)^2▒|𝑥^3−𝑥| 𝑑𝑥 |𝑥^3−𝑥|=|𝑥(𝑥^2−1)| =|𝑥| |𝑥^2−1| =|𝑥| |(𝑥−1)(𝑥+1)| =|𝑥| |𝑥−1||𝑥+1| Thus, 𝑥=0, 𝑥=1 & 𝑥=−1 ∴ |𝑥^3−𝑥|= {█((−𝑥)(−(𝑥−1))(−(𝑥+1)) 𝑖𝑓 𝑥<−1@(−𝑥)"(−(𝑥−1))" (𝑥+1) 𝑖𝑓 −1≤𝑥<0@(𝑥)"(−(𝑥−1))" (𝑥+1) 𝑖𝑓 0≤𝑥<1@(𝑥)(𝑥−1)(𝑥+1) 𝑖𝑓 𝑥≥1)┤ |𝑥^3−𝑥|= {█(−𝑥(𝑥−1)(𝑥+1) 𝑖𝑓 𝑥<−1@𝑥(𝑥−1)(𝑥+1) 𝑖𝑓 −1≤𝑥<0@−𝑥(𝑥−1)(𝑥+1) 𝑖𝑓 0≤𝑥<1@𝑥(𝑥−1)(𝑥+1) 𝑖𝑓 𝑥≥1)┤ |𝑥^3−𝑥|= {█(−(𝑥^3−𝑥) 𝑖𝑓 𝑥<−1@(𝑥^3−𝑥) 𝑖𝑓 −1≤𝑥<0@−(𝑥^3−𝑥) 𝑖𝑓 0≤𝑥<1@(𝑥^3−𝑥) 𝑖𝑓 𝑥≥1)┤ Thus, our integration becomes ∫_(−1)^2▒|𝑥^3−𝑥| 𝑑𝑥 =∫_(−1)^0▒(𝑥^3−𝑥) 𝑑𝑥−∫_0^1▒(𝑥^3−𝑥) 𝑑𝑥+∫_1^2▒(𝑥^3−𝑥) 𝑑𝑥 =∫_(−1)^0▒𝑥^3 𝑑𝑥−∫_(−1)^0▒𝑥 𝑑𝑥−∫_0^1▒𝑥^3 𝑑𝑥+∫_0^1▒𝑥 𝑑𝑥+∫_1^2▒𝑥^3 𝑑𝑥−∫_1^2▒𝑥 𝑑𝑥 =[𝑥^4/4]_(−1)^0−[𝑥^2/2]_(−1)^0−[𝑥^4/4]_0^1+[𝑥^2/2]_0^1+[𝑥^4/4]_1^2−[𝑥^2/2]_1^2 =[(0 − (−1)^4)/4]−[(0 − (−1)^2)/2]−[((1)^4 − 0)/4]+[((1)^2 − 0)/2]+[((2)^4 − (1)^4)/4]−[((2)^2 − (1)^2)/2] =(−1)/4 − [(−1)/2] − 1/4 + 1/2 + [(16 − 1)/4]−[(4 − 1)/2] =(−1)/4 + 1/2 − 1/4 + 1/2 +15/4 − 3/2 =(−1 + 2 − 1 + 2 + 15 − 6)/4 =𝟏𝟏/𝟒
