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Left Riemann sum for f(x) = x^3 on [-2,2] with n=4

Consider the function f(x)=x3f(x) = x^3 on the interval [−2,2][-2, 2].

Partition [−2,2][-2, 2] into n=4n = 4 equal subintervals and use the left endpoint of each subinterval to compute the left Riemann sum approximation for ∫−22x3 dx\displaystyle\int_{-2}^{2} x^3 \, dx.

What is the value of the left Riemann sum L4L_4?

Hint

First find Δx=b−an\Delta x = \frac{b-a}{n}, then list the left endpoints x0,x1,x2,x3x_0, x_1, x_2, x_3 of the four subintervals. Compute ff at each left endpoint, add the results, and multiply the total by Δx\Delta x.

Step-by-step solution

Since a=−2a=-2, b=2b=2, and n=4n=4, the width of each subinterval is Δx=2−(−2)4=1\Delta x = \frac{2-(-2)}{4} = 1. The subintervals are [−2,−1],[−1,0],[0,1],[1,2][-2,-1], [-1,0], [0,1], [1,2], so the left endpoints are x0=−2x_0=-2, x1=−1x_1=-1, x2=0x_2=0, x3=1x_3=1.

Evaluating f(x)=x3f(x) = x^3 at each left endpoint gives f(−2)=−8f(-2) = -8, f(−1)=−1f(-1) = -1, f(0)=0f(0) = 0, and f(1)=1f(1) = 1.

The left Riemann sum is L4=Δx∑i=03f(xi)=1⋅(−8+(−1)+0+1)=−8L_4 = \Delta x \sum_{i=0}^{3} f(x_i) = 1 \cdot (-8 + (-1) + 0 + 1) = -8.

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