Question Number 115229 by mathdave last updated on 24/Sep/20

find the mean value of   y=(5/(2−x−3x^2 ))  between x=−(1/3) and  x=(1/3)

Answered by Olaf last updated on 24/Sep/20

  y^_  = (1/((1/3)−(−(1/3))))∫_(−(1/3)) ^(+(1/3)) (5/(2−x−3x^2 ))dx  y^_  = −(5/2)∫_(−(1/3)) ^(+(1/3)) (1/(x^2 +(1/3)x−(2/3)))dx  y^_  = −(5/2)∫_(−(1/3)) ^(+(1/3)) (1/((x+(1/6))^2 −((25)/(36))))dx  y^_  = −(5/2)∫_(−(1/3)) ^(+(1/3)) (1/(((25)/(36))[((6/5)(x+(1/6)))^2 −1]))dx  u = (1/5)(6x+1)  y^_  = −3∫_(−(1/5)) ^(+(3/5)) (1/(u^2 −1))du  y^_  = −3[(1/2)ln∣((u−1)/(u+1))∣]_(−(1/5)) ^(3/5)   y^_  = −(3/2)[ln∣((−(2/5))/(8/5))∣−ln∣((−(6/5))/(4/5))∣]  y^_  = −(3/2)[ln((1/4))−ln((3/2))]  y^_  = −(3/2)ln((1/6))  y^_  = (3/2)ln6

Commented bymathdave last updated on 24/Sep/20

gud work

Answered by Bird last updated on 24/Sep/20

the mean value is (1/(b−a))∫_a ^b f(x)dx  =(1/((1/3)−(−(1/3))))∫_(−(1/3)) ^(1/(3 ))  (5/(−3x^2 −x+2))dx  =−((15)/2) ∫_(−(1/3)) ^(1/3)  (dx/(3x^2 +x−2))  Δ =1−4(−6) =25 ⇒x_1 =((−1+5)/6)  =(2/3) and x_2 =((−1−5)/6) =−1 ⇒  M_f  =−((15)/2) ∫_(−(1/3)) ^(1/3)  (dx/(3(x−(2/3))(x+1)))  =−(5/2)×(3/5) ∫_(−(1/3)) ^(1/3)  ((1/(x−(2/3)))−(1/(x+1)))dx  =−(3/2)[ln∣((x−(2/3))/(x+1))∣]_(−(1/3)) ^(1/3)   =−(3/2){ln∣((−(1/3))/(4/3))∣−ln∣((−1)/(2/3))∣}  =−(3/2){ln((3/4))−ln((3/2))}  =−(3/2)ln((3/4)×(2/3)) =−(3/2)ln((1/2))  =((3ln(2))/2)