Question

For what $a<0$ does the inequality ${\int }_a^0\left(3^{-2x}{2.3}^{-x}\right)dx⩾0$ hold true?

Answer 1

${\int }_a^0{3}^{-2x}2.3^{-x}dx=2{\int }_a^0{3}^{-3x}dx$

Apply substituion: $-3x=u$

$-3dx=du$

$dx=\frac{-1}{3}du$

$=2{\int }_{-3a}^0{3}^u\left(\frac{-1}{3}\right)du$

$=\frac{-2}{3}{\int }_{-3a}^0{3}^{u}du=\frac{-2}{3}\left[\frac{3^u}{\ln 3}{|}_{-3a}^0\right]$

$=\frac{-2}{3}[\frac{1}{\ln 3}-\frac{3^{-3a}}{\ln 3}]\ge 0$

$\Rightarrow [1-3^{-3a}]\le 0$

For this term to be ${}^{\prime }\le 0^{\prime }$

$1-3^{-3a}\le 0$

$1\le 3^{-3a}\Rightarrow 3^{3a}\le 1$

if $-1<a<0$ then $3a\ge 1$ and then $3^{3a}\text{/}\le 1$

Therefore, $a\le -1$.