Let $f(x) = e^x$ then $\int\limits_{-1}^{1}f(x)dx=\int\limits_{-1}^{1}e^xdx$ a = -1, b = 1, nh = b - a = 1 + 1 = 2 Now, $f(a)=f(-1)=e^{-1}$ f (a + h) = f (-1 + h) = $e^{-1+h}$ f (a + 2h) = f (-1 + 2h) = $e^{-1+2h}$ . . . . . . . . .. . . . . . . . .. . . . . . . . . $f(a+(n-1)h)$ $=f(-1+(n-1)h)$ $=e^{-1+(n-1)h}$ By definition $=\lim\limits_{h\to 0}h[e^{-1}(1+e^h+e^{2h}+\dots+e^{(n-1)h})]$$=\lim\limits_{h\to 0}h\cdot\frac{1}{e}\cdot\frac{1\cdot(e^{nh}-1)}{e^h-1}$$=\frac{1}{e}\cdot\frac{e^2-1}{\log e}=e-\frac{1}{e}$