Question

How large should \(n\) be to guarantee that the Simpson's Rule approximation to \(\int_0^1 {{e^{{x^2}}}} dx\)is accurate to within\(0.00001\)?

Short answer

The value of \(n\) should be 20.

Step-by-step solution

Definition:

Simpson's Rule is used for approximating the integral of a function between two limits\(a\,\& \,b\).It's supported on knowing the area under a parabola, or a plane curve.

Differentiate\(f(x)\):

Given integral is\(\int_0^1 {{e^{{x^2}}}} dx\).

Let \(f(x) = {e^{{x^2}}}\)

Differentiating \({f^\prime }(x) = 2x{e^{{x^2}}}\)

Now differentiate\(f'(x)\)to get:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) = 2x{e^{{x^2}}}(2x) + 2{e^{{x^2}}}\\ = 4{x^2}{e^{{x^2}}} + 2{e^{{x^2}}}\end{aligned}\)

Step 3: \(nth\,\)derivative:

The 3rd derivative of\(f(x)\)is :

\({f^{\prime \prime \prime }}(x) = \left( {4{x^2} + 2} \right)(2x){e^{{x^2}}} + 8x{e^{{x^2}}}\)

\( = \left( {8{x^3} + 4x + 8x} \right){e^{{x^2}}}\)

And then

\(\begin{aligned}{c}{f^{(4)}}(x) = \left( {8{x^3} + 12x} \right)(2x){e^{{x^2}}} + \left( {24{x^2} + 12} \right){e^{{x^2}}}\\ = \left( {16{x^4} + 24{x^2} + 24{x^2} + 12} \right){e^{{x^2}}}\end{aligned}\)

Determine bound for \({f^4}(x)\):

As \({f^{(4)}}(x)\) is an increasing function for \({\rm{x}} > 0\).

Therefore, \(\left| {{f^4}(x)} \right| = \left( {16{x^4} + 48{x^2} + 12} \right){e^{{x^2}}} \le (16 + 48 + 12)e\)

Or \(\left| {{f^4}(x)} \right| \le 76e\,for\,0 \le x \le 1\)

Let\(k = 76e,a = 0,b = 1\)

Determine value of \(n\):

For getting the error \(0.00001\) in Simpson's rule

The value of \({\rm{n}}\) should be such that:

\(\begin{aligned}{l}\frac{{k{{(b - a)}^5}}}{{180{n^4}}} < 0.00001\\ \Rightarrow \frac{{76e}}{{180{n^4}}} < 0.00001\\ \Rightarrow {n^4} > \frac{{76e}}{{0.0018}}\\ \Rightarrow n > \sqrt(4){{\frac{{76e}}{{0.0018}}}} \approx 18.4\end{aligned}\)

Next even number after 18 is 20

So the value is \(n = 20\) .