Question
Estimate the area under the graph of
f(x) = 3 cos(x)
from
x = 0
to
x =
𝜋
2
using four approximating rectangles and right endpoints, sketch the graph and the rectangles, and determine if your estimate is an underestimate or an overestimate.
Answers
GPT-4o mini
To estimate the area under the graph of \( f(x) = 3 \cos(x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) using four approximating rectangles with right endpoints, follow these steps:
### Step 1: Calculate the width of each rectangle
The width \( \Delta x \) of each rectangle is given by:
\[
\Delta x = \frac{b - a}{n}
\]
where \( a = 0 \), \( b = \frac{\pi}{2} \), and \( n = 4 \).
Calculating this:
\[
\Delta x = \frac{\frac{\pi}{2} - 0}{4} = \frac{\pi}{8}
\]
### Step 2: Determine the right endpoints
The right endpoints \( x_i \) for \( i = 1, 2, 3, 4 \) are calculated as:
- For \( i = 1 \): \( x_1 = 0 + \Delta x = 0 + \frac{\pi}{8} = \frac{\pi}{8} \)
- For \( i = 2 \): \( x_2 = \frac{\pi}{8} + \Delta x = \frac{\pi}{8} + \frac{\pi}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \)
- For \( i = 3 \): \( x_3 = \frac{2\pi}{8} + \Delta x = \frac{3\pi}{8} \)
- For \( i = 4 \): \( x_4 = \frac{3\pi}{8} + \Delta x = \frac{4\pi}{8} = \frac{\pi}{2} \)
The right endpoints are:
- \( x_1 = \frac{\pi}{8} \)
- \( x_2 = \frac{\pi}{4} \)
- \( x_3 = \frac{3\pi}{8} \)
- \( x_4 = \frac{\pi}{2} \)
### Step 3: Evaluate the function at each right endpoint
Calculating \( f(x_i) \):
- \( f(x_1) = 3 \cos\left(\frac{\pi}{8}\right) \)
- \( f(x_2) = 3 \cos\left(\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \)
- \( f(x_3) = 3 \cos\left(\frac{3\pi}{8}\right) \)
- \( f(x_4) = 3 \cos\left(\frac{\pi}{2}\right) = 3 \cdot 0 = 0 \)
### Step 4: Estimate the area using right endpoints
Now, we can estimate the area \( A \) using the formula:
\[
A \approx \sum_{i=1}^{n} f(x_i) \Delta x
\]
Calculating the individual areas:
\[
A \approx (3 \cos\left(\frac{\pi}{8}\right) + \frac{3\sqrt{2}}{2} + 3 \cos\left(\frac{3\pi}{8}\right) + 0) \cdot \frac{\pi}{8}
\]
Using cosine values:
- \( \cos\left(\frac{\pi}{8}\right) \approx 0.92388 \)
- \( \cos\left(\frac{3\pi}{8}\right) \approx 0.38268 \)
Plugging in the values:
\[
f(x_1) \approx 3 \cdot 0.92388 \approx 2.77164
\]
\[
f(x_3) \approx 3 \cdot 0.38268 \approx 1.14804
\]
Now substituting these values:
\[
A \approx (2.77164 + 2.12132 + 1.14804 + 0) \cdot \frac{\pi}{8}
\]
Calculating the sum:
\[
A \approx (6.04100) \cdot \frac{\pi}{8} \approx \frac{6.04100 \pi}{8} \approx 0.75513 \pi \approx 2.369 \text{ (approximately)}
\]
### Conclusion: Underestimate or Overestimate?
Since \( f(x) = 3 \cos(x) \) is decreasing on the interval \( [0, \frac{\pi}{2}] \), using right endpoints will yield an overestimate. Thus, our estimate of the area using right endpoints with four rectangles is an overestimate.
### Sketch
To sketch the graph:
1. Draw the curve for \( f(x) = 3 \cos(x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \).
2. Draw the rectangles starting from \( x_1 \) to \( x_2 \), \( x_2 \) to \( x_3 \), \( x_3 \) to \( x_4 \).
3. The height of each rectangle is given by \( f(x_i) \) where \( x_i \) is the right endpoint.
Illustratively, the graph should show a smooth decrease from \( f(0) = 3 \) to \( f\left(\frac{\pi}{2}\right) = 0\), as well as the rectangles extending to the values found from the evaluation at the right endpoints.
### Step 1: Calculate the width of each rectangle
The width \( \Delta x \) of each rectangle is given by:
\[
\Delta x = \frac{b - a}{n}
\]
where \( a = 0 \), \( b = \frac{\pi}{2} \), and \( n = 4 \).
Calculating this:
\[
\Delta x = \frac{\frac{\pi}{2} - 0}{4} = \frac{\pi}{8}
\]
### Step 2: Determine the right endpoints
The right endpoints \( x_i \) for \( i = 1, 2, 3, 4 \) are calculated as:
- For \( i = 1 \): \( x_1 = 0 + \Delta x = 0 + \frac{\pi}{8} = \frac{\pi}{8} \)
- For \( i = 2 \): \( x_2 = \frac{\pi}{8} + \Delta x = \frac{\pi}{8} + \frac{\pi}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \)
- For \( i = 3 \): \( x_3 = \frac{2\pi}{8} + \Delta x = \frac{3\pi}{8} \)
- For \( i = 4 \): \( x_4 = \frac{3\pi}{8} + \Delta x = \frac{4\pi}{8} = \frac{\pi}{2} \)
The right endpoints are:
- \( x_1 = \frac{\pi}{8} \)
- \( x_2 = \frac{\pi}{4} \)
- \( x_3 = \frac{3\pi}{8} \)
- \( x_4 = \frac{\pi}{2} \)
### Step 3: Evaluate the function at each right endpoint
Calculating \( f(x_i) \):
- \( f(x_1) = 3 \cos\left(\frac{\pi}{8}\right) \)
- \( f(x_2) = 3 \cos\left(\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \)
- \( f(x_3) = 3 \cos\left(\frac{3\pi}{8}\right) \)
- \( f(x_4) = 3 \cos\left(\frac{\pi}{2}\right) = 3 \cdot 0 = 0 \)
### Step 4: Estimate the area using right endpoints
Now, we can estimate the area \( A \) using the formula:
\[
A \approx \sum_{i=1}^{n} f(x_i) \Delta x
\]
Calculating the individual areas:
\[
A \approx (3 \cos\left(\frac{\pi}{8}\right) + \frac{3\sqrt{2}}{2} + 3 \cos\left(\frac{3\pi}{8}\right) + 0) \cdot \frac{\pi}{8}
\]
Using cosine values:
- \( \cos\left(\frac{\pi}{8}\right) \approx 0.92388 \)
- \( \cos\left(\frac{3\pi}{8}\right) \approx 0.38268 \)
Plugging in the values:
\[
f(x_1) \approx 3 \cdot 0.92388 \approx 2.77164
\]
\[
f(x_3) \approx 3 \cdot 0.38268 \approx 1.14804
\]
Now substituting these values:
\[
A \approx (2.77164 + 2.12132 + 1.14804 + 0) \cdot \frac{\pi}{8}
\]
Calculating the sum:
\[
A \approx (6.04100) \cdot \frac{\pi}{8} \approx \frac{6.04100 \pi}{8} \approx 0.75513 \pi \approx 2.369 \text{ (approximately)}
\]
### Conclusion: Underestimate or Overestimate?
Since \( f(x) = 3 \cos(x) \) is decreasing on the interval \( [0, \frac{\pi}{2}] \), using right endpoints will yield an overestimate. Thus, our estimate of the area using right endpoints with four rectangles is an overestimate.
### Sketch
To sketch the graph:
1. Draw the curve for \( f(x) = 3 \cos(x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \).
2. Draw the rectangles starting from \( x_1 \) to \( x_2 \), \( x_2 \) to \( x_3 \), \( x_3 \) to \( x_4 \).
3. The height of each rectangle is given by \( f(x_i) \) where \( x_i \) is the right endpoint.
Illustratively, the graph should show a smooth decrease from \( f(0) = 3 \) to \( f\left(\frac{\pi}{2}\right) = 0\), as well as the rectangles extending to the values found from the evaluation at the right endpoints.
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