Question

The value of ${\int }_0^1\frac{2^{2x+1}-5^{2x-1}}{{10}^x}dx$ is

• A. $\frac{3}{5}\left[\frac{2}{{\log }_e(\frac{2}{5})}+\frac{1}{2{\log }_e(\frac{5}{2})}\right]$
• B. $-\frac{3}{5}\left[\frac{2}{{\log }_e(\frac{2}{5})}+\frac{1}{2{\log }_e(\frac{5}{2})}\right]$
• C. $\frac{3}{5}\left[\frac{2}{{\log }_e(\frac{2}{5})}-\frac{1}{2{\log }_e(\frac{5}{2})}\right]$
• D. $\frac{1}{2}[\log 2-\log 3]$

Answer 1

Answer: (B)

$-\frac{3}{5}\left[\frac{2}{{\log }_e(\frac{2}{5})}+\frac{1}{2{\log }_e(\frac{5}{2})}\right]$

${\int }_0^1\left(\frac{2^{2x+1}+5^{2x-1}}{10x}\right)dx$
$2{\int }_0^1{\left(\frac{4}{10}\right)}^{xdx}-1/5{\int }_0^1{\frac{25}{10}}^{x}dx$
$=2\frac{(\frac{4}{10}{)}^x}{log\frac{\left(4\right)}{10}}{\int }_0^1-\frac{\frac{1}{5}\left(\frac{25}{10}\right)}{log\left(\frac{25}{10}\right)}$
$=\frac{2\left[\right(\frac{4}{10})-1]}{log\left(2/5\right)}-\frac{1/5\left[\right(\frac{25}{10}-1\left)\right]}{log\left(5/2\right)}$
$\frac{2(-3/5)}{l0g\left(2/5\right)}-\frac{1/5\left(3/2\right)}{log\left(5/2\right)}$
$=-3/5[\frac{2}{log\left(2/5\right)}+\frac{1}{2log}(5/2\left)\right].$