Recall that the error in using the Trapezoidal rule to estimate f(x) dr is M (b – a)3 |Er| < 12n2 where n is the number of subintervals and M is the maximum of f" on [a, b]. Determine the smallest number of subintervals n so that the error in estimating 2 4 4 dr 3 3 is less that 10

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### Understanding the Trapezoidal Rule Error Estimation

Recall that the error in using the Trapezoidal rule to estimate the definite integral \(\int_{a}^{b} f(x) \, dx\) is given by:

\[ |E_T| \leq \frac{M(b - a)^3}{12n^2} \]

where:
- \( n \) is the number of subintervals,
- \( M \) is the maximum value of the second derivative \( f''(x) \) over the interval \([a, b]\).

### Problem Statement

Determine the smallest number of subintervals \( n \) so that the error in estimating the integral:

\[ \int_{0}^{1} \left( \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \right) \, dx \]

is less than \( 10^{-3} \).

### Detailed Explanation

The given integral is:

\[ \int_{0}^{1} \left( \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \right) \, dx \]

To apply the trapezoidal rule, we need to consider the second derivative of the integrand:

\[ f(x) = \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \]

Calculating the first and second derivatives:

\[ f'(x) = \frac{4}{3} x^3 + 2 x^2 + 1 \]
\[ f''(x) = 4 x^2 + 4 x \]

To find \( M \), which is the maximum value of \( f''(x) \) on the interval \([0, 1]\):

\[ f''(x) = 4 x^2 + 4 x \]

Since \( f''(x) \) is a quadratic function that increases on \([0, 1]\), it reaches its maximum at \( x = 1 \).

\[ f''(1) = 4(1)^2 + 4(1) = 8 \]

Thus, \( M = 8 \).

Substituting back into the
Transcribed Image Text:### Understanding the Trapezoidal Rule Error Estimation Recall that the error in using the Trapezoidal rule to estimate the definite integral \(\int_{a}^{b} f(x) \, dx\) is given by: \[ |E_T| \leq \frac{M(b - a)^3}{12n^2} \] where: - \( n \) is the number of subintervals, - \( M \) is the maximum value of the second derivative \( f''(x) \) over the interval \([a, b]\). ### Problem Statement Determine the smallest number of subintervals \( n \) so that the error in estimating the integral: \[ \int_{0}^{1} \left( \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \right) \, dx \] is less than \( 10^{-3} \). ### Detailed Explanation The given integral is: \[ \int_{0}^{1} \left( \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \right) \, dx \] To apply the trapezoidal rule, we need to consider the second derivative of the integrand: \[ f(x) = \frac{1}{3} x^4 + \frac{2}{3} x^3 + x + \frac{4}{3} \] Calculating the first and second derivatives: \[ f'(x) = \frac{4}{3} x^3 + 2 x^2 + 1 \] \[ f''(x) = 4 x^2 + 4 x \] To find \( M \), which is the maximum value of \( f''(x) \) on the interval \([0, 1]\): \[ f''(x) = 4 x^2 + 4 x \] Since \( f''(x) \) is a quadratic function that increases on \([0, 1]\), it reaches its maximum at \( x = 1 \). \[ f''(1) = 4(1)^2 + 4(1) = 8 \] Thus, \( M = 8 \). Substituting back into the
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