To estimate the area under the curve of $f(x) = 4x + 1$ from $x = 0$ to $x = 3$ using six rectangles and the right endpoint, follow these steps:

1. Determine the interval width: $\text{Interval width} = \frac{3 - 0}{6} = 0.5$

2. Set up the right endpoints: Since we're using six rectangles with a width of $0.5$, the right endpoints will be $x = 0.5, 1, 1.5, 2, 2.5,$ and $3$.

3. Calculate $f(x)$ at each right endpoint:

   • $f(0.5) = 4(0.5) + 1 = 2 + 1 = 3$
   • $f(1) = 4(1) + 1 = 4 + 1 = 5$
   • $f(1.5) = 4(1.5) + 1 = 6 + 1 = 7$
   • $f(2) = 4(2) + 1 = 8 + 1 = 9$
   • $f(2.5) = 4(2.5) + 1 = 10 + 1 = 11$
   • $f(3) = 4(3) + 1 = 12 + 1 = 13$

4. Calculate the area of each rectangle and add them up: Each rectangle's area is $f(x_i) \times 0.5$, where $f(x_i)$ is the function value at each right endpoint. $\text{Total area} = (3 \times 0.5) + (5 \times 0.5) + (7 \times 0.5) + (9 \times 0.5) + (11 \times 0.5) + (13 \times 0.5)$ $= (1.5 + 2.5 + 3.5 + 4.5 + 5.5 + 6.5) = 24$

So, the estimated area under the curve from $x = 0$ to $x = 3$ is approximately 24.00 square units.

Would you like further details or have any questions?

Related Questions:

1. How would the area estimate change if left endpoints were used instead?
2. What would happen if the number of rectangles increased to improve accuracy?
3. Can we calculate the exact area under $f(x) = 4x + 1$ from $x = 0$ to $x = 3$ using integration?
4. How would this process differ if $f(x)$ were a nonlinear function?
5. What is the significance of using rectangles in estimating areas under curves?

Tip:

When estimating areas, increasing the number of rectangles generally increases accuracy because the approximation better conforms to the curve’s shape.