Left Riemann sum for f(x) = 2^x on [0,3] with n = 6

The width of each subinterval is $\Delta x = \frac{3-0}{6} = 0.5$. The subinterval endpoints are $x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2, x_5 = 2.5, x_6 = 3$. For a left Riemann sum, we evaluate $f$ at the left endpoint of each subinterval, namely $x_0, x_1, x_2, x_3, x_4, x_5$.

The left sum is $L_6 = \Delta x \left[f(x_0)+f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)\right] = 0.5\left[2^0 + 2^{0.5} + 2^1 + 2^{1.5} + 2^2 + 2^{2.5}\right]$. Computing each term: $2^0 = 1$, $2^{0.5} \approx 1.4142$, $2^1 = 2$, $2^{1.5} \approx 2.8284$, $2^2 = 4$, $2^{2.5} \approx 5.6569$. Summing these gives approximately $16.8995$, so $L_6 \approx 0.5 \times 16.8995 \approx 8.4497$.