We need a maximum enlargement of n = 18 terms in order to guarantee this level of accuracy for Simpon's Rule.
The estimation of Simpson's rule follows an approach where the value of the definite integral [tex]\int\limits^a_b f{x} \, dx[/tex] is approximated by using a parabolic method of approximation for each subinterval of the interval [a, b].
The formula for calculating Simpon's rule can be expressed as:
[tex]\mathbf{\int ^b_a f(x) d \simeq \dfrac{\Delta x}{3}[y_o+4y_1+2y_2+4y_3+2y_4+...+4{y_{n-1} +y_n}}[/tex]
Given that:
[tex]\mathbf{\int ^1_0 7ex^2 dx }[/tex],
We need to determine the fourth derivative; i.e.
[tex]\mathbf{f'(x) = 14xe^{x^2}}[/tex]
[tex]\mathbf{f''(x) = 14ex^2 +28x^2e^{x^2} = (14+28x^2)e^{x^2}}[/tex]
[tex]\mathbf{f'''(x) = 56xe^{x^2} +2(14+28x^2)e^{x^2} = (84x+48x^3)e^{x^2}}[/tex]
[tex]\mathbf{f^{iv}(x) = (84+144x^2)e^{x^2} +2x(84x+48x^3)e^{x^2} = (84x+312x^2+96x^4)e^{x^2}}[/tex]
The maximum value of the 4th derivative on the interval [0,1] is:
[tex]\mathbf{f^{iv}(1) = 624e}[/tex] and the error is bounded by:
[tex]\mathbf{\Bigg| \dfrac{f^{iv}(1) (1-0)^5}{180n^4} \Bigg| = \dfrac{624e}{180n^4}=\dfrac{52e}{15n^4}}[/tex]
Now for the error to be less than 0.0001; we have:
[tex]\mathbf{\dfrac{52e}{15n^4} < 0.0001}[/tex]
[tex]\mathbf{52e < 0.0015n^4}[/tex]
[tex]\mathbf{n^4 > \dfrac{52e}{0.0015}}[/tex]
[tex]\mathbf{n > \sqrt[4]{ \dfrac{52e}{0.0015}}}[/tex]
n ≅ 17.52.
Therefore, we can conclude that we need a maximum of n = 18 terms in order to guarantee this level of accuracy for Simpon's Rule.
Learn more about Simpson's Rule here:
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