Question Number 33867 by Joel578 last updated on 26/Apr/18
$$\mathrm{Given}\:{A}\left({t}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{between}\: \\ $$$${y}\:=\:{x}^{\mathrm{2}} \:+\:{tx}\:\mathrm{and}\:\mathrm{x}−\mathrm{axis},\:\:\mathrm{0}\:<\:{t}\:<\:\mathrm{2} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{propability}\:\mathrm{we}\:\mathrm{choose}\:{t} \\ $$$$\mathrm{so}\:\frac{\mathrm{1}}{\mathrm{48}}\:\leqslant\:{A}\left({t}\right)\:\leqslant\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Answered by MJS last updated on 26/Apr/18
![x^2 +tx=0 x(x+t)=0 x_1 =0 x_2 =−t ∣∫_(−t) ^0 (x^2 +tx)dx∣=∣[(x^3 /3)+((tx^2 )/2)]_(−t) ^0 ∣=∣(t^3 /3)−(t^3 /2)∣=(t^3 /6) (1/(48))≤(t^3 /6)≤(1/(16)) (1/8)≤t^3 ≤(3/8) (1/2)≤t≤((3)^(1/3) /2) ((((3)^(1/3) /2)−(1/2))/2)=(((3)^(1/3) −1)/4)≈.1106 ⇒ 11.06%](https://www.tinkutara.com/question/Q33872.png)
$${x}^{\mathrm{2}} +{tx}=\mathrm{0} \\ $$$${x}\left({x}+{t}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0} \\ $$$${x}_{\mathrm{2}} =−{t} \\ $$$$\mid\underset{−{t}} {\overset{\mathrm{0}} {\int}}\left({x}^{\mathrm{2}} +{tx}\right){dx}\mid=\mid\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{tx}^{\mathrm{2}} }{\mathrm{2}}\underset{−{t}} {\overset{\mathrm{0}} {\right]}}\mid=\mid\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{3}} }{\mathrm{2}}\mid=\frac{{t}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{48}}\leqslant\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\leqslant\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\leqslant{t}^{\mathrm{3}} \leqslant\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\leqslant{t}\leqslant\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\mathrm{4}}\approx.\mathrm{1106}\:\Rightarrow\:\mathrm{11}.\mathrm{06\%} \\ $$
Commented by Joel578 last updated on 26/Apr/18
$${thank}\:{you}\:{very}\:{much} \\ $$