Given-a-b-x-2-3x-x-3-dx-11-2-where-a-lt-3-lt-b-a-2b-8-Find-a-b-x-dx-

Question Number 132463 by EDWIN88 last updated on 14/Feb/21

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 {\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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