Question

Evaluate ${\int }_0^1{e}^{2-3x}dx$ as a limit of a sum.

Answer 1

$Letf\left(x\right)=e^{2-3x},then{\int }_0^1{e}^{2-3x}dx={\int }_0^{1}f\left(x\right)dxWeknowthat{\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{lim}h\left[f\right(a)+f(a+h)+f(a+2h)+....+f\{a+(n-1)h\left\}\right]where,h=\frac{b-a}{n}Here,a=0,b=1andnh=1andf\left(x\right)=e^{2-3x}\therefore f\left(0\right)=e^2;f(0+h)=e^{2-3h}andsoon.\therefore {\int }_0^1{e}^{2-3x}dx=\underset{h\to 0}{lim}h\left\{f\right(0)+f(0+h)+f(0+2h)+...+f(0+(n-1)h\left)\right\}=\underset{h\to 0}{lim}h\{e^2+e^{2-3h}+e^{2-6h}+....+e^{2-3(n-1)h}\}=\underset{h\to 0}{lim}h{e}^2\{1+e^{-3h}+e^{-6h}+....e^{-3(n-1)h}\}=\underset{h\to 0}{lim}\frac{h{e}^2\{1-(e^{-3h}{)}^n\}}{1-e^{-3h}}(thisisaGPseries,wherea=1,r=|e^{-3h}|<1)=\underset{h\to 0}{lim}\frac{e^2\{1-(e^{-3\left(1\right)}\left)\right\}}{\frac{1-e^{-3h}}{-3h}(-3)}[\because nh=1]=\frac{e^2\{1-(e^{-3}\left)\right\}}{3\underset{h\to 0}{lim}\frac{e^{-3h}-1}{-3h}}=\frac{1}{3}(e^2-e^{-1})(\because \underset{x\to 0}{lim}\frac{e^x-1}{x}=1)$