Given ∫_a ^( b) ((x^2 −3x)/(∣x−3∣)) dx = ((11)/2) where { ((a<3<b)),((a+2b=8)) :} Find ∫


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Question Number 132463 by EDWIN88 last updated on 14/Feb/21
$Given ∫_a ^( b) ((x^2 −3x)/(∣x−3∣)) dx = ((11)/2) where { ((a<3<b)),((a+2b=8)) :} Find ∫_a ^b ∣x∣ dx.$
$Given \int_{a}^{b} \frac{x^{2} - 3 x}{∣ x - 3 ∣} dx = \frac{11}{2} where {\begin{cases} a < 3 < b \\ a + 2 b = 8 \end{cases}$ $Find \int_{a}^{b} ∣ x ∣ dx .$
	Answered by bemath last updated on 14/Feb/21
	$From a<3<b we get { ((a−3<0 and)),((b−3>0)) :} the integral becomes ∫_a ^3 −((x(x−3))/(x−3))dx+∫_3 ^b ((x(x−3))/(x−3)) dx=((11)/2) (1/(2 ))[ x^2 ]_3 ^b −(1/2) [ x^2 ]_a ^3 = ((11)/2) ⇔ b^2 −9−(9−a^2 )=11 ⇒b^2 +a^2 = 29 ∧ a=8−2b ⇒b^2 +4b^2 −32b+64−29=0 ⇔5b^2 −32b+35=0 ⇔(5b−7)(b−5)=0 → { ((b=(7/5) (rejected))),((b=5 (accept))) :} we get a=−2 . So ∫^( 5) _(−2) ∣x∣ dx = ∫^( 0) _(−2) −x dx+∫_0 ^5 x dx = −(1/2)(−4)+(1/2)(25)=((29)/2)$
	$From a < 3 < b we get {\begin{cases} a - 3 < 0 a n d \\ b - 3 > 0 \end{cases}$ $the integral becomes \int_{a}^{3} - \frac{x (x - 3)}{x - 3} d x + \int_{3}^{b} \frac{x (x - 3)}{x - 3} d x = \frac{11}{2}$ $\frac{1}{2} {[x^{2}]}_{3}^{b} - \frac{1}{2} {[x^{2}]}_{a}^{3} = \frac{11}{2}$ $\Leftrightarrow b^{2} - 9 - (9 - a^{2}) = 11$ $\Rightarrow b^{2} + a^{2} = 29 \land a = 8 - 2 b$ $\Rightarrow b^{2} + 4 b^{2} - 32 b + 64 - 29 = 0$ $\Leftrightarrow 5 b^{2} - 32 b + 35 = 0$ $\Leftrightarrow (5 b - 7) (b - 5) = 0 \to {\begin{cases} b = \frac{7}{5} (rejected) \\ b = 5 (a c c e p t) \end{cases}$ $w e g e t a = - 2 . S o {\int_{- 2}}^{5} ∣ x ∣ dx =$ ${\int_{- 2}}^{0} - x dx + \int_{0}^{5} x dx = - \frac{1}{2} (- 4) + \frac{1}{2} (25) = \frac{29}{2}$
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