Question Number 132463 by EDWIN88 last updated on 14/Feb/21
$$\:\mathrm{Given}\:\int_{{a}} ^{\:{b}} \:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}}{\mid{x}−\mathrm{3}\mid}\:\mathrm{dx}\:=\:\frac{\mathrm{11}}{\mathrm{2}}\:\mathrm{where}\:\begin{cases}{{a}<\mathrm{3}<{b}}\\{{a}+\mathrm{2}{b}=\mathrm{8}}\end{cases} \\ $$$$\:\mathrm{Find}\:\int_{{a}} ^{{b}} \:\mid{x}\mid\:\mathrm{dx}.\: \\ $$
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)](https://www.tinkutara.com/question/Q132464.png)
$$\mathrm{From}\:{a}<\mathrm{3}<{b}\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{{a}−\mathrm{3}<\mathrm{0}\:{and}}\\{{b}−\mathrm{3}>\mathrm{0}}\end{cases} \\ $$$$\mathrm{the}\:\mathrm{integral}\:\mathrm{becomes}\:\int_{{a}} ^{\mathrm{3}} −\frac{{x}\left({x}−\mathrm{3}\right)}{{x}−\mathrm{3}}{dx}+\int_{\mathrm{3}} ^{{b}} \frac{{x}\left({x}−\mathrm{3}\right)}{{x}−\mathrm{3}}\:{dx}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}\:\:}\left[\:{x}^{\mathrm{2}} \:\right]_{\mathrm{3}} ^{{b}} −\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:{x}^{\mathrm{2}} \:\right]_{{a}} ^{\mathrm{3}} \:=\:\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:{b}^{\mathrm{2}} −\mathrm{9}−\left(\mathrm{9}−{a}^{\mathrm{2}} \right)=\mathrm{11} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \:=\:\mathrm{29}\:\wedge\:{a}=\mathrm{8}−\mathrm{2}{b} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} −\mathrm{32}{b}+\mathrm{64}−\mathrm{29}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{5}{b}^{\mathrm{2}} −\mathrm{32}{b}+\mathrm{35}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{5}{b}−\mathrm{7}\right)\left({b}−\mathrm{5}\right)=\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{b}=\frac{\mathrm{7}}{\mathrm{5}}\:\left(\mathrm{rejected}\right)}\\{{b}=\mathrm{5}\:\left({accept}\right)}\end{cases} \\ $$$${we}\:{get}\:{a}=−\mathrm{2}\:.\:{S}\mathrm{o}\:\underset{−\mathrm{2}} {\int}^{\:\:\:\mathrm{5}} \mid\mathrm{x}\mid\:\mathrm{dx}\:= \\ $$$$\underset{−\mathrm{2}} {\int}^{\:\:\:\mathrm{0}} −\mathrm{x}\:\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{5}} \mathrm{x}\:\mathrm{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{4}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{25}\right)=\frac{\mathrm{29}}{\mathrm{2}} \\ $$