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5.1 Approximating Areas under Curves

NOT on exam 3

How to find area under \( f(x) \) (\( f(x) > 0 \)) on \( a \le x \le b \)

For example, what is the area under \( f(x) = \sin x \) on \( [0, \pi/2] \)?

Graph of y = sin(x) from 0 to pi/2 with the area under the curve shaded in red and labeled area = ?

Not an elementary shape (triangle, rectangle, etc.) so we cannot use geometry.

Riemann Sum

One way to estimate the area is to use a Riemann Sum.

  • Divide region into rectangles and sum the areas.
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\( f(x) = \sin x \) on \( [0, \pi/2] \) using a Riemann Sum

First, decide how many rectangles to use.

As an example, let's use 4. So divide \( [0, \pi/2] \) into 4 parts. This gives us the grid points:

0, \( \pi/8 \), \( \pi/4 \), \( 3\pi/8 \), \( \pi/2 \)
Graph of sin(x) with 4 right-endpoint rectangles labeled 1 through 4 approximating the area.

Then we decide where on each subinterval to sample the rectangle weight/height.

Typical choices:

  • Right endpoint
  • Left endpoint
  • Midpoint

Here, as an example, let's use Right endpoints.

\( R_4 \) (four rectangles, right endpoint):

\[ R_4 = \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{8}\right) + \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{8}\right) \sin\left(\frac{3\pi}{8}\right) + \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{2}\right) = \boxed{1.183} \]

Note: Each term represents (width of rectangle) \( \times \) (height of rectangle) for rectangles 1, 2, 3, and 4 respectively.

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So we estimate area under \( f(x) = \sin x \) on \( [0, \pi/2] \) to be around 1.183.

We know this is an over estimation because each rectangle has part sticking out over the curve.

included in the 1.183 but shouldn't be

Graph of sin(x) from 0 to pi/2 with four right-endpoint rectangles showing overestimation areas shaded in red.
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Left Endpoint Approximation

Now let's try the left endpoints, still use 4 rectangles.

unaccounted for by rectangles

note â‘  has zero height

Graph of sin(x) from 0 to pi/2 with four left-endpoint rectangles. The first rectangle has zero height.
\[ L_4 = \left(\frac{\pi}{8}\right) \sin(0) + \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{8}\right) + \left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{8}\right) \sin\left(\frac{3\pi}{8}\right) = 0.791 \]

\( L_4 \): left end, 4 rectangles

\( (\pi/8) \): width of â‘ 

\( \sin(0) \): height of â‘ 

So, 0.791 must be an under estimate since there are regions that the rectangles cannot cover.

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Now we can confidently say, whatever the true area under \( f(x) = \sin(x) \) on \( [0, \frac{\pi}{2}] \) is, it must be no smaller than 0.791 and no greater than 1.183.

\[ 0.791 \le \text{true area} \le 1.183 \]

If we used the midpoint with 4 rectangles, we would get an estimate of 1.006.

In comparison, the true area (we'll learn how to find it later in Ch. 4) is exactly 1.

If \( f(x) \) is an increasing function on \( (a, b) \), then using Right endpoints results in an over estimate.

If \( f(x) \) is an increasing function on \( (a, b) \), then using Left endpoints results in an under estimate.

(The opposite is true if \( f(x) \) is decreasing)

Whatever endpoint we use \( (R, L, M) \), the more rectangles we use the better the estimate.

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Why?

Graph of an increasing curve with 2 right-endpoint rectangles showing large red-shaded excess areas.
Graph of the same curve with 4 right-endpoint rectangles showing smaller red-shaded excess areas.

We are including less of the excess by using more rectangles.

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In general, regardless of the choice of endpoints, we can write the approximation as

\[ A \approx f(x_1)\Delta x + f(x_2)\Delta x + \dots + f(x_n)\Delta x \]

Sample points (L, R, M)

These are the points used to evaluate the function for each rectangle.

\(\Delta x\) is the width of each rectangle

\[ \Delta x = \frac{b-a}{n} \]

upper limit

\[ = \sum_{k=1}^{n} f(x_k)\Delta x \]

lower limit

Summation (Sigma) notation

means we add up all terms starting at \(k=1\) and ending at \(k=n\) (\(k=1, 2, 3, \dots, n\))

For example,

\[ \sum_{k=1}^{5} k^2 = \underbrace{(1)^2}_{k=1} + \underbrace{(2)^2}_{k=2} + \underbrace{(3)^2}_{k=3} + \underbrace{(4)^2}_{k=4} + \underbrace{(5)^2}_{k=5} \]\[ = 1 + 4 + 9 + 16 + 25 = 55 \]

k=5: upper limit stop

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We also need to be able to write out sigma notation given the sum of terms

\[ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} \text{ in sigma notation} \]

[Note: The above sequence is crossed out in the original notes in favor of the following example]

\[ \underbrace{\frac{1}{3}}_{k=1} + \underbrace{\frac{1}{4}}_{k=2} + \underbrace{\frac{1}{5}}_{k=3} + \underbrace{\frac{1}{6}}_{k=4} + \underbrace{\frac{1}{7}}_{k=5} \text{ in sigma notation} \]
  • lower limit is 1
  • upper limit is 5

Relationship between k and each term

  • numerator is 1 regardless of k
  • denominator is 2 greater than k
\[ = \sum_{k=1}^{5} \frac{1}{k+2} \]