The integral [tex]\int\limits^1_0 {ex^2} \, dx[/tex] cannot be evaluated by finding an anti-derivative. Find a lower bound for the integer n to use for the Trapezoidal Rule in order to yield an error of no more than 10⁻⁶ for the approximation of the integral. You may want to use one of the following inequalities. If f(x) = ex², then f"(x) ≤ 6e and f ^{iv}(x) ≤ 76e for 0 ≤ x ≤ 1.

A trapezoid is a quadrilateral since n must be an integer, n= 369 is the smallest integer value of n that meets the stated accuracy criteria.

Trapezoidal rule:

Consider the integral,[tex]\int^{1}_{0} e^{x^{2}}\ dx.\\\\[/tex]

Here

[tex]\to \ f (x) = e^{x^2}\ , a=0 \ \ and \ \ b=1.\\\\[/tex]

Given:

[tex]\to f"(x) \leq 6e\ \ and\ \ f^{(iv)}\ (x) \leq 76e \ \ for\ \ 0\leq x \leq 1\\\\ So, K = 6e.[/tex]

In the Trapezoidal Rule, use the error estimate.

[tex]\to |E_T|=\int^{b}_{a} f(x) dx-T_n-| \leq \frac{K(b-a)^3}{ 12n^2}\\\\[/tex]

The number of subintervals n should be chosen so that

[tex]\to |E_T| \leq 10^{-5}. \\\\[/tex]

[tex]\to \frac{K(b-a)^3}{12n^2}\leq 10^{-5}\\\\\to \frac{6e(1-0)^3}{12n^2} \leq \frac{1}{ 100000}\\\\\to \frac{e}{2n^2}\leq \frac{1}{ 100000}\\\\\to \frac{2n^2}{e}\geq 100000\\\\ \to n^2 \geq 100000 \times \frac{e}{2}\\\\ \to n^2 \geq 50000 \times e\\\\\to n\geq \sqrt{50000 \times e}\approx 368.665\\\\[/tex]

Because n must be an integer [tex]n= 369[/tex] is the smallest integer value of n that meets the stated accuracy criteria.

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