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Midpoint sum in sigma notation and evaluation

Consider approximating the area under f(x)=xf(x) = \sqrt{x} on the interval [1,4][1,4] using a midpoint sum with n=6n = 6 rectangles of equal width.

Partition [1,4][1,4] into n=6n = 6 equal subintervals to approximate ∫14x dx\int_1^4 \sqrt{x}\,dx using the midpoint rule.

First, determine Δx\Delta x and identify the midpoint xˉi\bar{x}_i of the ii-th subinterval in terms of ii. Then write the midpoint sum as a single expression in sigma notation (a sum over ii from 11 to 66), and evaluate it to obtain a numerical approximation for the area.

What is the value of the midpoint sum (written in sigma notation) that approximates the area under f(x)=xf(x)=\sqrt{x} on [1,4][1,4]?

Hint

First find Δx=(4−1)/6\Delta x = (4-1)/6, then write the general midpoint xˉi=1+Δx(i−12)\bar{x}_i = 1 + \Delta x(i - \tfrac12) for i=1,…,6i=1,\dots,6. Plug this into f(xˉi)f(\bar{x}_i) and multiply by Δx\Delta x, then sum over ii; you can evaluate the six terms numerically rather than using a closed-form formula, since x\sqrt{x} has no simple sum formula.

Step-by-step solution

With a=1a=1, b=4b=4, n=6n=6, the width is Δx=4−16=12\Delta x = \frac{4-1}{6} = \frac{1}{2}. The ii-th subinterval is [1+12(i−1), 1+12i][1+\frac{1}{2}(i-1),\, 1+\frac{1}{2}i], so its midpoint is xˉi=1+12(i−12)=34+i2\bar{x}_i = 1 + \frac{1}{2}\left(i - \frac{1}{2}\right) = \frac{3}{4} + \frac{i}{2} for i=1,…,6i = 1,\dots,6. The midpoint sum in sigma notation is M6=∑i=16Δx f(xˉi)=∑i=161234+i2.M_6 = \sum_{i=1}^{6} \Delta x \, f(\bar{x}_i) = \sum_{i=1}^{6} \frac{1}{2}\sqrt{\frac{3}{4}+\frac{i}{2}}.

Now evaluate the six terms: the midpoints are xˉ1=1.25, xˉ2=1.75, xˉ3=2.25, xˉ4=2.75, xˉ5=3.25, xˉ6=3.75\bar{x}_1=1.25,\ \bar{x}_2=1.75,\ \bar{x}_3=2.25,\ \bar{x}_4=2.75,\ \bar{x}_5=3.25,\ \bar{x}_6=3.75. Computing 12xˉi\frac12\sqrt{\bar{x}_i} for each and adding gives approximately 0.559+0.662+0.750+0.829+0.901+0.968≈4.6690.559+0.662+0.750+0.829+0.901+0.968 \approx 4.669. So M6=∑i=161234+i2≈4.67M_6 = \displaystyle\sum_{i=1}^{6}\frac{1}{2}\sqrt{\frac{3}{4}+\frac{i}{2}} \approx 4.67.

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