Here $a = 0,\; b = 1$ , $f(x) = e^{2-3x}$ ; $h = \frac{1-0}{n} = \frac{1}{n}$ $\Rightarrow nh = 1$ $f(a + h) = f(0 + h)$ $= e^{2-3h}$ $f(a + 2h) = f(0 + 2h)$ $= e^{2-6h}$ ................................................................................ $f(a + (n-1)h) = f(0 + (n-1)h)$ $= e^{2-3(n-1)h}$$= \lim\limits_{h \to 0} h \left[ e^{2} \times \left( \frac{1 \cdot (e^{-3hn} - 1)}{e^{-3h} - 1} \right) \right]$$= \lim\limits_{h \to 0} \left[ h e^{2} \times \frac{e^{-3} - 1}{e^{-3h} - 1} \right]$$= \frac{e^{2} - (1 - e^{-3})}{ \frac{e^{-3h} - 1}{-3h} \cdot (-3) }$$= \frac{e^{2} \left(1 - \frac{1}{e^{3}}\right)}{3} = \frac{1}{3} \left(e^{2} - \frac{1}{e}\right)$