We are tasked with estimating the upper bound of the absolute error for approximating the integral $\int_0^1 \sin(x^2) \, dx$ using 4 subintervals. We are also asked to find how large $n$ (the number of subintervals) must be for the approximation to be accurate within 0.001.

Step 1: Understanding the Trapezoidal Rule and Error Estimation

To approximate the integral using the Trapezoidal Rule, we divide the interval $[0,1]$ into $n$ subintervals and apply the rule:

$T_n = \frac{\Delta x}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right)$ where $\Delta x = \frac{1-0}{n} = \frac{1}{n}$, and $f(x) = \sin(x^2)$.

The error bound for the Trapezoidal Rule is given by:

$|E_T| \leq \frac{(b-a)^3}{12n^2} \max_{x \in [a,b]} |f''(x)|$

For the interval $[0,1]$, this simplifies to:

$|E_T| \leq \frac{1}{12n^2} \max_{x \in [0,1]} |f''(x)|$

Step 2: Calculating the Second Derivative

We first need to compute the second derivative of $f(x) = \sin(x^2)$.

1. First derivative: $f'(x) = 2x \cos(x^2)$

2. Second derivative: $f''(x) = 2\cos(x^2) - 4x^2 \sin(x^2)$

Now, we need to estimate the maximum value of $|f''(x)|$ on $[0,1]$.

Step 3: Estimating $\max_{x \in [0,1]} |f''(x)|$

To get an upper bound for $|f''(x)|$, observe that:

• $\cos(x^2)$ oscillates between -1 and 1, so $2\cos(x^2)$ has a maximum value of 2.
• $\sin(x^2)$ oscillates between -1 and 1, and $-4x^2 \sin(x^2)$ is maximized at $x=1$, where it equals 4.

Thus, a rough upper bound for $|f''(x)|$ is $2 + 4 = 6$. This is a conservative estimate of the maximum of $|f''(x)|$.

Step 4: Calculating the Error Bound for 4 Subintervals

Now, using $n = 4$ subintervals, the error bound becomes:

$|E_T| \leq \frac{1}{12(4)^2} \cdot 6 = \frac{1}{12 \cdot 16} \cdot 6 = \frac{6}{192} = 0.03125$

So, the upper bound for the absolute error when using 4 subintervals is 0.031250.

Step 5: Finding $n$ for Accuracy within 0.001

We need to find $n$ such that the error is less than 0.001:

$\frac{1}{12n^2} \cdot 6 \leq 0.001$

This simplifies to:

[ \frac{6}{12n^2} \leq 0.001 \quad \Rightarrow