Math Problem Statement

How large should n be to guarantee that the Simpson's Rule approximation to 1 ∫ 0 7e^x^2 dx is accurate to within 0.00001? n ≥

Solution

To determine the minimum number of subintervals, nn, required for Simpson's Rule to approximate the integral

01ex2dx\int_0^1 e^{x^2} \, dx

accurately to within 0.000010.00001, we need to use the error bound for Simpson's Rule. The error bound for Simpson’s Rule is given by the formula:

ESM(ba)5180n4E_S \leq \frac{M(b - a)^5}{180 n^4}

where:

  • ESE_S is the error of Simpson’s rule,
  • MM is the maximum value of the fourth derivative of the integrand f(x)=ex2f(x) = e^{x^2} over the interval [a,b][a, b],
  • nn is the number of subintervals (which must be even),
  • a=0a = 0, and b=1b = 1 are the limits of integration.

We are asked to find nn such that the error is less than or equal to 0.000010.00001:

ES0.00001E_S \leq 0.00001

Step 1: Find the fourth derivative of f(x)=ex2f(x) = e^{x^2}

The function f(x)=ex2f(x) = e^{x^2} has the following derivatives:

  • First derivative: f(x)=2xex2f'(x) = 2x e^{x^2}
  • Second derivative: f(x)=(4x2+2)ex2f''(x) = (4x^2 + 2) e^{x^2}
  • Third derivative: f(3)(x)=(8x3+12x)ex2f^{(3)}(x) = (8x^3 + 12x) e^{x^2}
  • Fourth derivative: f(4)(x)=(16x4+48x2+12)ex2f^{(4)}(x) = (16x^4 + 48x^2 + 12) e^{x^2}

We need to find the maximum value of f(4)(x)f^{(4)}(x) over the interval [0,1][0, 1].

Step 2: Estimate the maximum value of the fourth derivative on [0,1][0, 1]

To find the maximum, we evaluate f(4)(x)f^{(4)}(x) at the endpoints x=0x = 0 and x=1x = 1:

  • At x=0x = 0, f(4)(0)=12f^{(4)}(0) = 12.
  • At x=1x = 1, f(4)(1)=(16+48+12)e1=76ef^{(4)}(1) = (16 + 48 + 12)e^1 = 76e.

Thus, the maximum value of f(4)(x)f^{(4)}(x) on the interval [0,1][0, 1] is approximately:

M76e205.2M \approx 76e \approx 205.2

Step 3: Solve for nn

Now, we substitute MM, a=0a = 0, b=1b = 1, and ES0.00001E_S \leq 0.00001 into the error bound formula:

0.00001205.2×(10)5180n40.00001 \geq \frac{205.2 \times (1 - 0)^5}{180 n^4}

Simplifying:

0.00001205.2180n40.00001 \geq \frac{205.2}{180 n^4}

n4205.2180×0.00001n^4 \geq \frac{205.2}{180 \times 0.00001}

n4205.20.0018114000n^4 \geq \frac{205.2}{0.0018} \approx 114000

n114000434.4n \geq \sqrt[4]{114000} \approx 34.4

Since nn must be even, the smallest integer nn is n=36n = 36.

Final Answer:

To guarantee that the Simpson's Rule approximation is accurate to within 0.000010.00001, you need at least n36n \geq 36 subintervals.


Would you like more details on any part of the solution? Or perhaps explore another question?

Here are 5 related questions:

  1. What is the role of the fourth derivative in determining the error bound for Simpson's Rule?
  2. How do higher derivatives affect the accuracy of numerical methods like Simpson’s Rule?
  3. Can Simpson's Rule be used for indefinite integrals?
  4. How does the error bound for Simpson's Rule compare to the one for Trapezoidal Rule?
  5. How can we improve the accuracy of Simpson's Rule without increasing nn?

Tip: Increasing nn significantly improves the accuracy of Simpson’s Rule, but be mindful of computational cost—larger nn values require more function evaluations.

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Math Problem Analysis

Mathematical Concepts

Simpson's Rule
Numerical Integration
Error Bounds
Derivatives
Approximation Methods

Formulas

Error Bound for Simpson's Rule: E_S ≤ (M(b - a)^5) / (180 n^4)
Fourth derivative of f(x) = e^{x^2}

Theorems

Simpson's Rule Error Bound Theorem

Suitable Grade Level

College Level (Calculus)