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The problem in the image asks to use the **Midpoint Rule** with \( n = 4 \) to approximate the area under the graph of \( f(x) = 3 \tan(x) \) over the interval \( [0, \frac{\pi}{4}] \). Here's how we can approach this:

### Steps to solve using the Midpoint Rule:
1. **Divide the interval** \( [0, \frac{\pi}{4}] \) into \( n = 4 \) subintervals. The width of each subinterval is:
   \[
   \Delta x = \frac{b - a}{n} = \frac{\frac{\pi}{4} - 0}{4} = \frac{\pi}{16}
   \]

2. **Find the midpoints** of each subinterval:
   The midpoints \( x_i^* \) for the subintervals are:
   - \( x_1^* = 0 + \frac{\Delta x}{2} = \frac{\pi}{32} \)
   - \( x_2^* = \frac{\pi}{16} + \frac{\Delta x}{2} = \frac{3\pi}{32} \)
   - \( x_3^* = \frac{2\pi}{16} + \frac{\Delta x}{2} = \frac{5\pi}{32} \)
   - \( x_4^* = \frac{3\pi}{16} + \frac{\Delta x}{2} = \frac{7\pi}{32} \)

3. **Evaluate the function** \( f(x) = 3 \tan(x) \) at these midpoints:
   \[
   f(x_1^*) = 3 \tan\left( \frac{\pi}{32} \right), \quad f(x_2^*) = 3 \tan\left( \frac{3\pi}{32} \right), \quad f(x_3^*) = 3 \tan\left( \frac{5\pi}{32} \right), \quad f(x_4^*) = 3 \tan\left( \frac{7\pi}{32} \right)
   \]

4. **Apply the Midpoint Rule formula**:
   \[
   A \approx \Delta x \left( f(x_1^*) + f(x_2^*) + f(x_3^*) + f(x_4^*) \right)
   \]
   where \( \Delta x = \frac{\pi}{16} \).

5. **Numerical calculation**:
   - \( f(x_1^*) = 3 \tan\left( \frac{\pi}{32} \right) \approx 3(0.09817) = 0.29451 \)
   - \( f(x_2^*) = 3 \tan\left( \frac{3\pi}{32} \right) \approx 3(0.29492) = 0.88476 \)
   - \( f(x_3^*) = 3 \tan\left( \frac{5\pi}{32} \right) \approx 3(0.49394) = 1.48182 \)
   - \( f(x_4^*) = 3 \tan\left( \frac{7\pi}{32} \right) \approx 3(0.69960) = 2.09880 \)

   Adding them:
   \[
   f(x_1^*) + f(x_2^*) + f(x_3^*) + f(x_4^*) = 0.29451 + 0.88476 + 1.48182 + 2.09880 = 4.75989
   \]

6. **Multiply by \( \Delta x \)** to get the approximate area:
   \[
   A \approx \frac{\pi}{16} \times 4.75989 \approx 0.19635 \times 4.75989 \approx 0.935
   \]

### Final Answer:
The approximate area under the curve using the Midpoint Rule is \( \boxed{0.935} \), rounded to three decimal places.

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Would you like more details on any step? Here are some related questions:

1. How does the Midpoint Rule compare to other numerical integration methods?
2. What are the advantages and limitations of using the Midpoint Rule?
3. Can we use more intervals to increase the accuracy of the Midpoint Rule?
4. How does the function \( f(x) = 3\tan(x) \) behave over different intervals?
5. What other real-world applications are there for the Midpoint Rule?

**Tip:** The accuracy of numerical integration methods increases as you use more subintervals, so always consider refining \( n \) if the estimate seems rough!

Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of the function f(x) = 3 tan(x) and the x-axis over the interval [0, π/4]. (Round your answer to three decimal places.)

Learn how to use the Midpoint Rule with n = 4 to approximate the area under the graph of f(x) = 3 tan(x) over the interval [0, π/4]. This detailed step-by-step solution covers key concepts in numerical integration, including how to calculate midpoints and apply the Midpoint Rule formula. Suitable for students in Grades 11-12.

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