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Right Riemann sum for f(x) = ln(x) on [1,4] with n=3

Consider the function f(x)=ln⁡(x)f(x) = \ln(x) on the interval [1,4][1, 4].

Partition [1,4][1,4] into n=3n = 3 subintervals of equal width, and use the right endpoint of each subinterval to compute the Riemann sum approximating ∫14ln⁡(x) dx\int_1^4 \ln(x)\,dx.

What is the value of the right Riemann sum R3R_3?

Hint

First find Δx=b−an\Delta x = \frac{b-a}{n}, then list the right endpoints x1,x2,x3x_1, x_2, x_3. The right sum is R3=Δx[f(x1)+f(x2)+f(x3)]R_3 = \Delta x \left[ f(x_1) + f(x_2) + f(x_3) \right].

Step-by-step solution

With a=1a=1, b=4b=4, n=3n=3, the width of each subinterval is Δx=4−13=1\Delta x = \frac{4-1}{3} = 1. The subintervals are [1,2],[2,3],[3,4][1,2], [2,3], [3,4], so the right endpoints are x1=2x_1 = 2, x2=3x_2 = 3, and x3=4x_3 = 4.

The right Riemann sum is R3=Δx[f(x1)+f(x2)+f(x3)]=1⋅[ln⁡(2)+ln⁡(3)+ln⁡(4)]R_3 = \Delta x\left[f(x_1) + f(x_2) + f(x_3)\right] = 1\cdot\left[\ln(2) + \ln(3) + \ln(4)\right]. Evaluating gives R3≈0.6931+1.0986+1.3863≈3.178R_3 \approx 0.6931 + 1.0986 + 1.3863 \approx 3.178.

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