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find-the-mean-value-of-y-5-2-x-3x-2-between-x-1-3-and-x-1-3-




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)
findthemeanvalueofy=52x3x2betweenx=13andx=13
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
y_=113(13)13+1352x3x2dxy_=5213+131x2+13x23dxy_=5213+131(x+16)22536dxy_=5213+1312536[(65(x+16))21]dxu=15(6x+1)y_=315+351u21duy_=3[12lnu1u+1]1535y_=32[ln2585ln6545]y_=32[ln(14)ln(32)]y_=32ln(16)y_=32ln6
Commented by mathdave last updated on 24/Sep/20
gud work
gudwork
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)
themeanvalueis1baabf(x)dx=113(13)131353x2x+2dx=1521313dx3x2+x2Δ=14(6)=25x1=1+56=23andx2=156=1Mf=1521313dx3(x23)(x+1)=52×351313(1x231x+1)dx=32[lnx23x+1]1313=32{ln1343ln123}=32{ln(34)ln(32)}=32ln(34×23)=32ln(12)=3ln(2)2

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