Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given integral is

[tex]\rm \: \displaystyle\int\limits_{0}^{2}\left(e^{x}-x\right) d x \\ \\ [/tex]

Here,

[tex]\rm \: a \: = \: 0 \\ [/tex]

[tex]\rm \: b \: = \: 2 \\ [/tex]

[tex]\rm \: f(x) \: = \: {e}^{x} - x \\ \\ [/tex]

Now,

[tex] \red{\rm \: nh \: = \: b - a \: = \: 2 - 0 = 2 }\\ \\ [/tex]

Now, Consider

[tex] \red{\rm \: f(x) \: = \: {e}^{x} - x} \\ [/tex]

Put x = a + rh = 0 + rh = rh

So, above expression can be rewritten as

[tex] \red{\rm \: f(rh) \: = \: {e}^{rh} - rh} \\ \\ [/tex]

Now, By definition of integral using limit as a sum, we have

[tex]\rm \: \displaystyle\int\limits_{a}^{b}f(x) d x = \lim_{h \to 0}h \: \displaystyle\sum_{r=0}^{n-1}\: f(a + rh) \\ \\ [/tex]

So, on substituting the values, we get

[tex]\rm \: \displaystyle\int \limits_{0}^{2}\left(e^{x}-x\right) d x \\ [/tex]

[tex]\rm \: = \: \displaystyle\lim_{h \to 0}h \: \displaystyle\sum_{r=0}^{n-1}\: ({e}^{rh} - rh) \\ [/tex]

[tex]\rm \: = \: \displaystyle\lim_{h \to 0}h\displaystyle\sum_{r=0}^{n-1}\: {e}^{rh} -\displaystyle\lim_{h \to 0}h\displaystyle\sum_{r=0}^{n-1}\: rh \\ [/tex]

[tex]\rm \: = \: \displaystyle\lim_{h \to 0}h\displaystyle(1 + {e}^{h} + {e}^{2h} + \cdots n \: terms) -\displaystyle\lim_{h \to 0} {h}^{2} \frac{n(n - 1)}{2} \\ [/tex]

Now, first term is a geometric progression with first term 1, common ratio [tex]{e}^{h} [/tex] and number of terms n.

We know,

Sum of n terms of a geometric progression having first term a, common ratio r and number of terms n is given by

[tex] \green{\boxed{ \bf{ \:S_n \: = \: \frac{a \: ( {r}^{n} - 1)}{r - 1} \: }}} \\ \\ [/tex]

So, using this result, we get

[tex]\rm \: = \: \displaystyle\lim_{h \to 0}h \times \dfrac{1( {e}^{nh} - 1)}{{e}^{h} - 1} -\displaystyle\lim_{h \to 0} \frac{nh(nh - h)}{2} \\ [/tex]

[tex]\rm \: = \: \displaystyle\lim_{h \to 0}h \times \dfrac{( {e}^{2} - 1)}{{e}^{h} - 1} -\displaystyle\lim_{h \to 0} \frac{2(2 - h)}{2} \\ [/tex]

[tex]\rm \: = \: ( {e}^{2} - 1) \displaystyle\lim_{h \to 0} \dfrac{h}{{e}^{h} - 1} -\displaystyle\lim_{h \to 0} (2 - h) \\ [/tex]

[tex]\rm \: = \: ( {e}^{2} - 1) \times 1 - (2 - 0) \\ [/tex]

[tex]\rm \: = \: {e}^{2} - 1 - 2 \\ [/tex]

[tex]\rm \: = \: {e}^{2} - 3 \\ \\ [/tex]

Hence,

[tex]\bf\implies \: \displaystyle\int\limits_{0}^{2}\left(e^{x}-x\right) d x \: = \: {e}^{2} - 3 \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]