Evaluate `int0 1e^(2-3x)dx`as a limit of a sum.

Explanation

`int_a^bf(x)dx=lim_(h to 0) h[f(a)+f(a+h)+f(a+2h)+-------+f(a+(n-1)h] `
where, `h=(b-a)/n`
`nh=1-0=1 `,br> ` implies I=int_0^1e^(2-3x)dx `
`=lim_(h to 0)h[f(0)+f(h)+f(2h)+ -----+f(n-1)h]`
`=lim _(h to 0)h[e^2+e^(2-3h)+e^(2-3(2h))+ ----+e^(2-3(n-1)h)]`
`=lim_(h to 0)he^2[1+e^-3h+e^-3(2h)+ -----+e^(-3(n-1)h)]`
`=lim _(h to 0) he^2{{(e^(-3h))^n-1}/[e^(-3h)-1]}`
`=lim_(h to 0) he^2 {{e^(-3nh)-1}/[e^(-3h)-1]}`
`=e^2 lim_(h to 0) (e^-3-1)/{[e^(-3h)-1]/(-3h)} times -1/3`
`implies I=e^2(e^(-3)-1) times -1/3`
`=1/3(e^2-e^(-1))`
Hence solved.