Approximating Areas: Fresh Take

Lumen Learning

Approximating Areas: Fresh Take

Estimate the area under a curve by adding up the areas of rectangles
Estimate the area under a curve using Riemann sums

Approximating Area

The Main Idea

Area under a curve can be approximated using rectangles
Two main methods: left-endpoint and right-endpoint approximations
Increasing the number of rectangles improves accuracy
Riemann sums formalize this approximation process
Partitions:
- Divide interval $[a,b]$ into n subintervals
- Regular partition: subintervals of equal width $\Delta x = \frac{(b-a)}{n}$
Left-Endpoint Approximation:
- $L_n = \sum_{i=1}^n f(x_{i-1})\Delta x$
- Rectangle heights based on function value at left endpoint
Right-Endpoint Approximation:
- $R_n = \sum_{i=1}^n f(x_i)\Delta x$
- Rectangle heights based on function value at right endpoint
Convergence:
- As $n$ increases, approximations generally become more accurate
- Left and right approximations often converge to the true area

Write in sigma notation and evaluate the sum of terms

2^{i}

$2^i$ for

i = 3, 4, 5, 6

$i=3,4,5,6$ .

$\displaystyle\sum_{i=3}^{6} 2^i=2^3+2^4+2^5+2^6=120$

Find the sum of the values of $f(x)=x^3$ over the integers $1,2,3,\cdots,10$ .

Using the formula, we have

\begin{matrix} 10 \sum i = 0 i^{3} & = \frac{(10)^{2} (10 + 1)^{2}}{4} = \frac{100 (121)}{4} = 3025 \end{matrix}

$\begin{array}{ll}\displaystyle\sum_{i=0}^{10} i^3 & =\frac{(10)^2(10+1)^2}{4} \\ & =\frac{100(121)}{4} \\ & =3025 \end{array}$

Find left and right-endpoint approximations for $\int_0^2 x^2 dx$ with $n = 4$ .

Partition:

$\Delta x = \frac{(2-0)}{4} = 0.5$

Subintervals:

$[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$

Left-Endpoint Approximation:

$L_4 = (0^2)(0.5) + (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5)$
$= 0 + 0.125 + 0.5 + 1.125 = 1.75$

Right-Endpoint Approximation:

$R_4 = (0.5^2)(0.5) + (1^2)(0.5) + (1.5^2)(0.5) + (2^2)(0.5)$
$= 0.125 + 0.5 + 1.125 + 2 = 3.75$

Comparison:

$L_4 = 1.75 < \text{ True Area } < R_4 = 3.75$

Note: The true value is $\int_0^2 x^2 dx = [\frac{1}{3}x^3]_0^2 = \frac{8}{3} \approx 2.67$

Riemann Sums

The Main Idea

Riemann sums generalize left and right endpoint approximations. They allow evaluation of the function at any point within each subinterval
As the number of subintervals increases, the approximation improves
Riemann sums can be used to define the area under a curve
Riemann Sum Definition:
- $\displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x$ where $x_i^*$ is any point in the subinterval $[x_{i-1},x_i]$
Area Under a Curve:
- $A=\underset{n\to \infty }{\lim}\displaystyle\sum_{i=1}^{n} f(x_i^*)\Delta x$
Upper and Lower Sums:
- Upper sum:
  - Choose $x_i^*$ for maximum $f(x_i^*)$ in each subinterval
- Lower sum:
  - Choose $x_i^*$ for minimum $f(x_i^*)$ in each subinterval
Convergence:
- For continuous functions, the limit of Riemann sums is unique
- The limit doesn’t depend on the choice of $x_i^*$

Find a lower sum for $f(x)= \sin x$ over the interval $[a,b]=[0,\frac{\pi }{2}]$ ; let $n=6$ .

Let’s first look at the graph in Figure 16 to get a better idea of the area of interest.

A graph of the function y = sin(x) from 0 to pi. It is set up for a left endpoint approximation from 0 to pi/2 and n=6. It is a lower sum. — Figure 16. The graph of $y= \sin x$ is divided into six regions: $\Delta x=\frac{\pi/2}{6}=\frac{\pi}{12}$ .

The intervals are $[0,\frac{\pi}{12}], \, [\frac{\pi}{12},\frac{\pi}{6}], \, [\frac{\pi}{6},\frac{\pi}{4}], \, [\frac{\pi}{4},\frac{\pi}{3}], \, [\frac{\pi}{3},\frac{5\pi}{12}]$ , and $[\frac{5\pi}{12},\frac{\pi}{2}]$ . Note that $f(x)= \sin x$ is increasing on the interval $[0,\frac{\pi}{2}]$ , so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum $\displaystyle\sum_{i=0}^{5} \sin x_i(\frac{\pi}{12})$ . We have

$\begin{array}{ll}A & \approx \sin (0)(\frac{\pi }{12})+ \sin (\frac{\pi }{12})(\frac{\pi }{12})+ \sin (\frac{\pi }{6})(\frac{\pi }{12})+ \sin (\frac{\pi }{4})(\frac{\pi }{12})+ \sin (\frac{\pi }{3})(\frac{\pi }{12})+ \sin (\frac{5\pi }{12})(\frac{\pi }{12}) \\ & =0.863 \end{array}$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “5.1 Approximating Areas” here (opens in new window).

Using the function $f(x)= \sin x$ over the interval $[0,\frac{\pi}{2}]$ , find an upper sum; let $n=6$ .

$A \approx 1.125$