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


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

$f i n d t h e m e a n v a l u e o f$ $y = \frac{5}{2 - x - 3 x^{2}} b e t w e e n x = - \frac{1}{3} a n d$ $x = \frac{1}{3}$
	Answered by Olaf last updated on 24/Sep/20

	$\overline{y} = \frac{1}{\frac{1}{3} - (- \frac{1}{3})} \int_{- \frac{1}{3}}^{+ \frac{1}{3}} \frac{5}{2 - x - 3 x^{2}} d x$ $\overline{y} = - \frac{5}{2} \int_{- \frac{1}{3}}^{+ \frac{1}{3}} \frac{1}{x^{2} + \frac{1}{3} x - \frac{2}{3}} d x$ $\overline{y} = - \frac{5}{2} \int_{- \frac{1}{3}}^{+ \frac{1}{3}} \frac{1}{{(x + \frac{1}{6})}^{2} - \frac{25}{36}} d x$ $\overline{y} = - \frac{5}{2} \int_{- \frac{1}{3}}^{+ \frac{1}{3}} \frac{1}{\frac{25}{36} [{(\frac{6}{5} (x + \frac{1}{6}))}^{2} - 1]} d x$ $u = \frac{1}{5} (6 x + 1)$ $\overline{y} = - 3 \int_{- \frac{1}{5}}^{+ \frac{3}{5}} \frac{1}{u^{2} - 1} d u$ $\overline{y} = - 3 {[\frac{1}{2} \ln ∣ \frac{u - 1}{u + 1} ∣]}_{- \frac{1}{5}}^{\frac{3}{5}}$ $\overline{y} = - \frac{3}{2} [\ln ∣ \frac{- \frac{2}{5}}{\frac{8}{5}} ∣ - \ln ∣ \frac{- \frac{6}{5}}{\frac{4}{5}} ∣]$ $\overline{y} = - \frac{3}{2} [\ln (\frac{1}{4}) - \ln (\frac{3}{2})]$ $\overline{y} = - \frac{3}{2} \ln (\frac{1}{6})$ $\overline{y} = \frac{3}{2} \ln 6$
	Commented by mathdave last updated on 24/Sep/20

	$g u d w o r k$
	Answered by Bird last updated on 24/Sep/20
	$the mean value is (1/(b−a))∫_a ^b f(x)dx =(1/((1/3)−(−(1/3))))∫_(−(1/3)) ^(1/(3 )) (5/(−3x^2 −x+2))dx =−((15)/2) ∫_(−(1/3)) ^(1/3) (dx/(3x^2 +x−2)) Δ =1−4(−6) =25 ⇒x_1 =((−1+5)/6) =(2/3) and x_2 =((−1−5)/6) =−1 ⇒ M_f =−((15)/2) ∫_(−(1/3)) ^(1/3) (dx/(3(x−(2/3))(x+1))) =−(5/2)×(3/5) ∫_(−(1/3)) ^(1/3) ((1/(x−(2/3)))−(1/(x+1)))dx =−(3/2)[ln∣((x−(2/3))/(x+1))∣]_(−(1/3)) ^(1/3) =−(3/2){ln∣((−(1/3))/(4/3))∣−ln∣((−1)/(2/3))∣} =−(3/2){ln((3/4))−ln((3/2))} =−(3/2)ln((3/4)×(2/3)) =−(3/2)ln((1/2)) =((3ln(2))/2)$
	$t h e m e a n v a l u e i s \frac{1}{b - a} \int_{a}^{b} f (x) d x$ $= \frac{1}{\frac{1}{3} - (- \frac{1}{3})} \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{5}{- 3 x^{2} - x + 2} d x$ $= - \frac{15}{2} \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{d x}{3 x^{2} + x - 2}$ $Δ = 1 - 4 (- 6) = 25 \Rightarrow x_{1} = \frac{- 1 + 5}{6}$ $= \frac{2}{3} a n d x_{2} = \frac{- 1 - 5}{6} = - 1 \Rightarrow$ $M_{f} = - \frac{15}{2} \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{d x}{3 (x - \frac{2}{3}) (x + 1)}$ $= - \frac{5}{2} \times \frac{3}{5} \int_{- \frac{1}{3}}^{\frac{1}{3}} (\frac{1}{x - \frac{2}{3}} - \frac{1}{x + 1}) d x$ $= - \frac{3}{2} {[l n ∣ \frac{x - \frac{2}{3}}{x + 1} ∣]}_{- \frac{1}{3}}^{\frac{1}{3}}$ $= - \frac{3}{2} {l n ∣ \frac{- \frac{1}{3}}{\frac{4}{3}} ∣ - l n ∣ \frac{- 1}{\frac{2}{3}} ∣}$ $= - \frac{3}{2} {l n (\frac{3}{4}) - l n (\frac{3}{2})}$ $= - \frac{3}{2} l n (\frac{3}{4} \times \frac{2}{3}) = - \frac{3}{2} l n (\frac{1}{2})$ $= \frac{3 l n (2)}{2}$
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