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Left-endpoint Riemann sum in sigma notation

Consider the function f(x)=exf(x) = e^x on the interval [0,2][0,2].

Partition [0,2][0,2] into n=4n=4 equal subintervals, and let L4L_4 denote the left-endpoint Riemann sum approximating ∫02ex dx\displaystyle\int_0^2 e^x\,dx. Determine Δx\Delta x and the left endpoints xi−1x_{i-1} of the subintervals, then write L4L_4 using sigma notation, expressed entirely in terms of the index ii.

Hint

First find Δx=(b−a)/n\Delta x = (b-a)/n, then express the left endpoint of the ii-th subinterval as xi−1=a+(i−1)Δxx_{i-1} = a + (i-1)\Delta x. Substitute this into f(xi−1)f(x_{i-1}) and sum over i=1i=1 to 44, multiplying each term by Δx\Delta x.

Step-by-step solution

Since a=0a=0, b=2b=2, and n=4n=4, the subinterval width is Δx=2−04=12\Delta x = \dfrac{2-0}{4} = \dfrac{1}{2}. The left endpoint of the ii-th subinterval is xi−1=0+(i−1)⋅12=i−12x_{i-1} = 0 + (i-1)\cdot \dfrac12 = \dfrac{i-1}{2}, for i=1,2,3,4i=1,2,3,4.

The left-endpoint Riemann sum is L4=∑i=14f(xi−1) ΔxL_4 = \sum_{i=1}^{4} f(x_{i-1})\,\Delta x. Substituting f(x)=exf(x) = e^x, Δx=12\Delta x = \frac12, and xi−1=i−12x_{i-1} = \frac{i-1}{2} gives L4=∑i=1412e(i−1)/2.L_4 = \sum_{i=1}^{4} \frac{1}{2} e^{(i-1)/2}. This is the desired sigma-notation expression for the left-endpoint sum.

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