question_answer

Evaluate \[\int_{0}^{1}{{{e}^{3x-2}}dx.}\]

Answer:

We have, \[\int_{a}^{b}{f(x)}\,dx=\underset{h\to 0}{\mathop{\lim }}\,h\,\,[f(a)+f(a+h)+...+f(a+(n-1)h)]\] Where,  \[h=\frac{b-a}{n}\]        Here, a = 2, b = 1 and \[f(x)={{e}^{3x-2}}\] Therefore, \[h=\frac{1}{n}\] \[\Rightarrow \] nh = 1 \[\therefore \,\,\,l=\int_{0}^{1}{{{e}^{3x-2}}dx}\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,h[f(0)+f(h)+f(2h)+...f(n-1)h]\] \[=\underset{h\to \,\,0}{\mathop{lim}}\,h[{{e}^{-2}}+{{e}^{3h-2}}+{{e}^{6h-2}}+...{{e}^{3(n-1)h-2}}]\] \[=\underset{h\to \,\,0}{\mathop{lim}}\,h{{e}^{-2}}[1+{{e}^{3h}}+{{e}^{3h(2)}}+...{{e}^{3(n-1)h}}]\] \[=\underset{h\to \,\,0}{\mathop{lim}}\,h{{e}^{-2}}\left[ \frac{{{({{e}^{3h}})}^{n}}-1}{{{e}^{3h}}-1} \right]\] \[=\underset{h\to \,\,0}{\mathop{lim}}\,h{{e}^{-2}}\left[ \frac{{{e}^{3h}}^{n}-1}{{{e}^{3h}}-1} \right]\] \[=\underset{h\to \,\,0}{\mathop{lim}}\,{{e}^{-2}}\left[ \frac{{{e}^{3}}-1}{\frac{{{e}^{3h}}-1}{2h}} \right]\times \frac{1}{3}\] \[={{e}^{-2}}({{e}^{3}}-1)\times \frac{1}{3}=\frac{1}{3}(e-{{e}^{-2}})\]