Question

Calculate the following integral:
${\int }_0^3[3^{1-x}+{\left(\frac{1}{3}\right)}^{2x-1}]dx$

Answer 1

${\int }_0^3{3}^{1-x}+{\left(\frac{1}{3}\right)}^{2x-1}dx=\underset{}{{\int }_0^3{3}^{1-x}dx}+\underset{}{{\int }_0^3{3}^{1-2x}dx}$

Apply substitution $u=1-x\Rightarrow -du=dx$ (for first integration)

$u=1-2x\Rightarrow du=-2dx\Rightarrow \frac{-1}{2}du=dx$(for second integration)

$\Rightarrow -{\int }_{+1}^{-2}{3}^{u}du+\left(\frac{-1}{2}\right){\int }_1^{-5}{3}^{u}du$

$\Rightarrow \frac{3^u}{\ln \left(3\right)}{|}_{-2}^1+\frac{1}{2}\left(\frac{3^u}{\ln 3}\right){|}_{-5}^1=\frac{3}{\ln 3}-\frac{3^{-2}}{\ln \left(3\right)}+\frac{1}{2}[\frac{3}{\ln 3}-\frac{3^{-5}}{\ln 3}]$

$=\frac{1}{\ln 3}[3-\frac{1}{9}+\frac{1}{2}[3-\frac{1}{3^5}]]$

$=\frac{1}{\ln 3}[\frac{26}{9}+\frac{1}{2}\left(\frac{728}{243}\right)]$

$=\frac{1066}{\ln \left(3\right)\times 243}$.