Question

Evaluate the following integrals as limit of a sum:

`int _0^2 e^x * dx`

Answer 1

Let f(x) = e^x, for 0 ≤ x ≤ 2

DIvide the closed interval [0, 2] into n equal subintervals each of length h at the points

0, 0 + h, 0 + 2h, ..., 0 + rh, ... 0 + nh = 2

i.e. 0,h, 2h, ..., rh, ..., nh = 2

∴ `h = 2/n` and n → ∞, h → 0

Here, a = 0

∴ f(a + rh) = f(0 + rh) = f(rh) = e^rh 

`because int_a^b f(x) dx = lim_(n -> infty) sum_(r = 1)^n f (a + rh)h`

`therefore int_0^2 e^x dx = lim_(r = 1)^n e^(rh) * h`

= `lim_(h -> 0) [h sum_(r = 1)^n e^(rh)]`    ...[as n → ∞, h → 0 ]

Now, `sum_(r  1)^n e^(rh) = e^h + e^(2h) + ... + e^(nh)`

= `(e^h [(e^h)^n - 1])/(e^h - 1)`

= `(e^h [e^(nh) - 1])/(e^h - 1)`

= `(e^h * (e^2 - 1))/(e^h - 1)    ...[because h = 2/n therefore nh = 2]`

= `(e^2 - 1) e^h/(e^b - 1)`

`therefore int_0^2 e^x dx = lim_(h -> 0) (h(e^2 - 1) e^h)/(e^h - 1)`

= `(e^2 - 1) lim_(h -> 0)(e^h/((e^h - 1)/h))`

= `(e^2 - 1) (lim_(h -> 0)e^h)/(lim_(h -> 0) ((e^h - 1)/h))`

= `(e^2 - 1)(lim_(h -> 0) e^h)/(lim _(h -> 0) ((e^h - 1)/h))`

= `(e^2 - 1) * e^0/1    ...[because lim_(h -> 0) (e^(h - 1))/h = 1]`

= e² − 1