PIL import Image import pytesseract

Load the image to extract text

image_path = "/mnt/data/1732898641189ln7ivqs3.jpg" extracted_text = pytesseract.image_to_string(Image.open(image_path))

extracted_textTo approximate the integral $\int_0^2 (4x^3 + x) dx$ using the midpoint rule with $n = 4$ subintervals, follow these steps:

Step 1: Calculate $\Delta x$

The interval $[0, 2]$ is divided into 4 subintervals: $\Delta x = \frac{2 - 0}{4} = 0.5$

Step 2: Determine midpoints

The subintervals are: [ [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2] ] The midpoints are: $x_1 = 0.25, \, x_2 = 0.75, \, x_3 = 1.25, \, x_4 = 1.75$

Step 3: Evaluate the function $f(x) = 4x^3 + x$ at the midpoints

$f(0.25) = 4(0.25)^3 + 0.25 = 4(0.015625) + 0.25 = 0.0625 + 0.25 = 0.3125$ $f(0.75) = 4(0.75)^3 + 0.75 = 4(0.421875) + 0.75 = 1.6875 + 0.75 = 2.4375$ $f(1.25) = 4(1.25)^3 + 1.25 = 4(1.953125) + 1.25 = 7.8125 + 1.25 = 9.0625$ $f(1.75) = 4(1.75)^3 + 1.75 = 4(5.359375) + 1.75 = 21.4375 + 1.75 = 23.1875$

Step 4: Compute the midpoint Riemann sum

The formula is: $\text{Midpoint Riemann Sum} = \Delta x \sum_{i=1}^n f(x_i)$ $\text{Midpoint Riemann Sum} = 0.5 \left[f(0.25) + f(0.75) + f(1.25) + f(1.75)\right]$ $\text{Midpoint Riemann Sum} = 0.5 \left[0.3125 + 2.4375 + 9.0625 + 23.1875\right]$ $\text{Midpoint Riemann Sum} = 0.5 \cdot 35 = 17.5$

Final Answer:

The approximation of the integral using the midpoint rule is: $\boxed{17.5}$

Would you like further explanation or a step-by-step breakdown of any part?

Related Questions:

1. How is the midpoint rule different from the left or right endpoint Riemann sums?
2. Can this result be verified by evaluating the exact integral?
3. What happens if we increase $n$ to 8 subintervals? How does it affect accuracy?
4. How can this method be extended to irregular intervals?
5. How does the midpoint rule compare to the trapezoidal rule?

Tip:

The midpoint rule often provides a better approximation than the left or right Riemann sums because it reduces error by sampling near the "center" of the interval.