The problem you provided asks to find the approximate area under the curve $y = 4x^2$ between $x = 1$ and $x = 3$ using the method of inscribed rectangles. The task specifies two different numbers of subintervals:

(a) $n = 5$ subintervals with $\Delta x = 0.2$, and (b) $n = 10$ subintervals with $\Delta x = 0.4$.

To solve this:

1. Formula for the area using rectangles: The area of each rectangle is the width $\Delta x$ times the height of the function at the left endpoint of the interval (since we are finding the area of the inscribed rectangles). The formula for the approximation is: $A \approx \sum_{i=0}^{n-1} f(x_i) \Delta x$ where $x_i = 1 + i \Delta x$ (since $x = 1$ is the starting point).

2. Given:

   • The function $y = 4x^2$.
   • The interval is from $x = 1$ to $x = 3$, so $\Delta x = \frac{3 - 1}{n}$.

(a) For $n = 5$ subintervals:

• The width of each subinterval is $\Delta x = \frac{3 - 1}{5} = 0.4$.
• For the left endpoint, $x_0 = 1, x_1 = 1.4, x_2 = 1.8, x_3 = 2.2, x_4 = 2.6$.
• The height of each rectangle is found by evaluating $y = 4x^2$ at these values.

The total area is approximately: $A = f(1) \times 0.4 + f(1.4) \times 0.4 + f(1.8) \times 0.4 + f(2.2) \times 0.4 + f(2.6) \times 0.4$ where:

• $f(1) = 4(1)^2 = 4$
• $f(1.4) = 4(1.4)^2 = 4 \times 1.96 = 7.84$
• $f(1.8) = 4(1.8)^2 = 4 \times 3.24 = 12.96$
• $f(2.2) = 4(2.2)^2 = 4 \times 4.84 = 19.36$
• $f(2.6) = 4(2.6)^2 = 4 \times 6.76 = 27.04$

Now, calculate the total area: $A \approx (4 + 7.84 + 12.96 + 19.36 + 27.04) \times 0.4$ $A \approx 71.2 \times 0.4 = 28.48$

So, the approximate area under the curve for part (a) is 28.48.

Would you like further clarification or assistance on part (b)?