Question

Evaluate $\displaystyle\sf \int\limits_0^2 e^x\:dx$ as the limit of sums.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\sf \int\limits_0^2 {e}^{x}dx}$

Here,

$\rm :\longmapsto\:a = 0$

$\rm :\longmapsto\:b = 2$

$\rm :\longmapsto\:f(x) = {e}^{x}$

Now,

$\red{\rm :\longmapsto\:nh = b - a = 2 - 0 = 2 - - - (1)}$

Now,

$\green{\rm :\longmapsto\:f(x) = {e}^{x} }$

$\green{\rm :\longmapsto\:Put \: \: x = a + rh = 0 + rh = rh}$

So,

$\green{\rm :\longmapsto\:f(rh) = {e}^{rh} }$

Now, By definition using limit as sum, we have

$\purple{\rm :\longmapsto\:\displaystyle\ \int\limits_a^b \: f(x)dx = \displaystyle\lim_{h \to 0} \sum_{r = 0}^{n - 1} \: f(a + rh)}$

On substituting the values, we get

$\purple{\rm :\longmapsto\:\displaystyle\ \int\limits_0^2 \: {e}^{x} dx = \displaystyle\lim_{h \to 0} \: h \: \sum_{r = 0}^{n - 1} \: {e}^{rh} }$

$\purple{\rm \: = \: \: \displaystyle\lim_{h \to 0} \: h(1 + {e}^{h} + {e}^{2h} + {e}^{3h} + - - - + {e}^{(n - 1)h}}$

We know,

☆ Sum of n terms of GP series is given by

$\boxed{ \rm{ S_n = \frac{a( {r}^{n} - 1) }{r - 1}}}$

where,

a is first term of GP series

r is the common ratio.

So, using this formula, we get

$\purple{\rm \: = \: \: \displaystyle\lim_{h \to 0} \: h \: \frac{1( {e}^{nh} - 1)}{ {e}^{h} - 1} }$

$\purple{\rm \: = \: \: \displaystyle\lim_{h \to 0} \: \: \frac{( {e}^{2} - 1)}{ \dfrac{ {e}^{h} - 1}{h} }}$

We know

$\boxed{ \rm{ \displaystyle\lim_{h \to 0} \frac{ {e}^{h} - 1}{h} = 1}}$

So, using this we get,

$\purple{\rm \: = \: \: ( {e}^{2} - 1)}$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\sf \int\limits_0^2 {e}^{x}dx = {e}^{2} - 1}$