Answer
$a)$ $\int_{{\,10}}^{{\,20}}$ $\frac{1}{x}$ $dx$ = $\int_{{\,1}}^{{\,2}}$ $\frac{1}{x}$ $dx$ = $\ln2$
$b)$ $\int_{{\,ka}}^{{\,kb}}$ $\frac{1}{x}$ $dx$ = $\int_{{\,a}}^{{\,b}}$ $\frac{1}{x}$ $dx$ = $\ln(\frac{b}{a})$
Work Step by Step
$a)$
$A$ = $\int_{{\,10}}^{{\,20}}$ $\frac{1}{x}$ $dx$
$A$ = $\ln{x}$ $|_{{\,10}}^{{\,20}}$
$A$ = $\ln20-\ln10$ = $\ln2$
$A$ = $\int_{{\,1}}^{{\,2}}$ $\frac{1}{x}$ $dx$
$A$ = $\ln{x}$ $|_{{\,1}}^{{\,2}}$
$A$ = $\ln2-\ln1$ = $\ln2$
$b)$
$A$ = $\int_{{\,ka}}^{{\,kb}}$ $\frac{1}{x}$ $dx$
$A$ = $\ln{x}$ $|_{{\,ka}}^{{\,kb}}$
$A$ = $\ln{kb}-\ln{ka}$ = $\ln(\frac{kb}{ka})$ = $\ln(\frac{b}{a})$
$A$ = $\int_{{\,a}}^{{\,b}}$ $\frac{1}{x}$ $dx$
$A$ = $\ln{x}$ $|_{{\,a}}^{{\,b}}$
$A$ = $\ln{b}-\ln{a}$ = $\ln(\frac{b}{a})$
