Question

${\int }_0^1{e}^{2-3x}dx$

Answer 1

${\int }_0^1{e}^{2-3x}\cdot dx$ $[\because \int e^x\cdot dx=e^x+c]$

$={\left[\frac{e^{2-3x}}{(2-3x{)}^{\prime }}\right]}_0^1={\left[\frac{e^{2-3x}}{-3}\right]}_0^1$

$=\frac{-1}{3}[e^{2-3x}{]}_0^1=\frac{-1}{3}[e^{2-3}-e^{2-0}]$

$=\frac{-1}{3}[e^{-1}-e^2]=\frac{-1}{3}[\frac{1}{e}-e^2]$.