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Left-endpoint Riemann sum in sigma notation

Consider the function f(x)=4−x2f(x) = 4 - x^2 on the interval [−1,2][-1, 2].

Partition [−1,2][-1, 2] into n=5n = 5 equal subintervals, and consider the left-endpoint Riemann sum approximating ∫−12(4−x2) dx\int_{-1}^{2} (4 - x^2)\, dx. Determine Δx\Delta x and the left endpoint xi−1x_{i-1} in terms of ii, then write the left-endpoint Riemann sum using sigma notation, with all numerical values filled in (do not evaluate the sum).

Hint

First compute Δx=b−an\Delta x = \frac{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 multiply by Δx\Delta x, summing from i=1i=1 to i=ni=n.

Step-by-step solution

The width of each subinterval is Δx=2−(−1)5=35\Delta x = \frac{2 - (-1)}{5} = \frac{3}{5}. The left endpoint of the ii-th subinterval is xi−1=−1+(i−1)⋅35x_{i-1} = -1 + (i-1)\cdot \frac{3}{5}, since we start at a=−1a = -1 and step forward by Δx\Delta x for each of the previous i−1i-1 subintervals.

The left-endpoint Riemann sum is ∑i=15f(xi−1) Δx\sum_{i=1}^{5} f(x_{i-1})\, \Delta x, where f(xi−1)=4−(−1+35(i−1))2f(x_{i-1}) = 4 - \left(-1 + \frac{3}{5}(i-1)\right)^2. Substituting Δx=35\Delta x = \frac{3}{5} gives the final sigma notation expression.

Number of terms in the Riemann sum

Hint
Step-by-step solution

The index i in the sigma notation starts at one, since the sum runs from i equals one to i equals five. This corresponds to the first subinterval, whose left endpoint is x sub zero, labeled using index i equals one. Recognizing this confirms the sum has exactly five terms, matching the five subintervals from partitioning the interval.

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