Ex 7.8, 3 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Ex 7.8
Ex 7.8, 1
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Transcript
Ex 7.8, 3 ∫_1^2▒〖(4𝑥3−5𝑥2+6𝑥+9) 𝑑𝑥〗 Step 1 :- F(𝑥)=∫1▒(4𝑥^3−5𝑥^2+6𝑥+9)𝑑𝑥 =(4𝑥^(3+1))/(3+1)−(5𝑥^(2+1))/(2+1)+(6𝑥^(1+1))/(1+1)+9𝑥 =(4𝑥^4)/4−(5𝑥^3)/3+(6𝑥^2)/2+9𝑥 = 𝑥^4−5/3 𝑥^3+〖3𝑥〗^2+9𝑥 Hence f(x)〖=𝑥〗^4−(3𝑥^3)/3+3𝑥^2+9𝑥 Step 2 :- ∫_1^2▒〖(4𝑥^2−9𝑥^2+6𝑥+9)𝑑𝑥 〗 =𝐹(2)−𝐹(1) =(2)^4−5/3 (2)^3+3(2)^2+9(2)−(1^4−5/3 ×1^3+3 ×1^2+9 ×1) =16−5/3 ×8+3 ×4+9 ×2−1+5/3−3−9 =16−40/3 +12+18−1+5/3−3−9 =33−35/3 =(3(33) − 35)/3 = 𝟔𝟒/𝟑
