To estimate the area under the curve $f(x) = \frac{1}{3}x^2 + \frac{1}{10}x^2 - \frac{6}{5}x + 4$ on the interval $[-3, 2]$ using 5 midpoint rectangles, we'll go through the following steps:

1. Simplify the Function

First, let's combine the like terms in the function to simplify it: $f(x) = \left(\frac{1}{3} + \frac{1}{10}\right)x^2 - \frac{6}{5}x + 4$ $f(x) = \frac{10 + 3}{30}x^2 - \frac{6}{5}x + 4$ $f(x) = \frac{13}{30}x^2 - \frac{6}{5}x + 4$

2. Divide the Interval into 5 Subintervals

The interval $[-3, 2]$ has a length of $2 - (-3) = 5$. Dividing this into 5 equal subintervals gives each subinterval a width of $\Delta x = \frac{5}{5} = 1$.

So, the subintervals are: [ [-3, -2], [-2, -1], [-1, 0], [0, 1], [1, 2] ]

3. Determine the Midpoints of Each Subinterval

The midpoints of these intervals are: $x_1 = -2.5, \quad x_2 = -1.5, \quad x_3 = -0.5, \quad x_4 = 0.5, \quad x_5 = 1.5$

4. Calculate the Function Values at the Midpoints

We need to calculate $f(x)$ at each of these midpoints:

$f(-2.5) = \frac{13}{30}(-2.5)^2 - \frac{6}{5}(-2.5) + 4$ $f(-1.5) = \frac{13}{30}(-1.5)^2 - \frac{6}{5}(-1.5) + 4$ $f(-0.5) = \frac{13}{30}(-0.5)^2 - \frac{6}{5}(-0.5) + 4$ $f(0.5) = \frac{13}{30}(0.5)^2 - \frac{6}{5}(0.5) + 4$ $f(1.5) = \frac{13}{30}(1.5)^2 - \frac{6}{5}(1.5) + 4$

5. Calculate the Area of Each Rectangle and Sum Them Up

Each rectangle has a width of $\Delta x = 1$, and the height of each rectangle is $f(x_i)$ for the corresponding midpoint $x_i$.

The total area estimate is: $\text{Total Area} = \sum_{i=1}^{5} f(x_i) \cdot \Delta x$

I will calculate these function values and the total area now.The function values at the midpoints are:

$f(-2.5) \approx 9.71, \quad f(-1.5) \approx 6.78, \quad f(-0.5) \approx 4.71, \quad f(0.5) \approx 3.51, \quad f(1.5) \approx 3.18$

The estimated total area under the curve using the 5 midpoint rectangles is:

$\text{Total Area} \approx 27.88$

If you need any more details or have further questions, feel free to ask!

Related Questions:

1. What is the actual area under the curve using integration?
2. How does the midpoint approximation compare to using left or right endpoints?
3. Can we approximate the area using trapezoidal or Simpson’s rule?
4. How would the area estimate change with more subintervals?
5. How can you determine the error in the midpoint approximation?
6. What are the applications of midpoint rule in real-life problems?
7. Can we visualize the midpoint approximation and compare it with other methods?
8. How does changing the interval bounds affect the area under the curve?

Tip:

When calculating areas using numerical methods, increasing the number of subintervals generally leads to more accurate results.