How large should n be to guarantee that the Simpson's Rule approximation on the Integral (from 0 to 1) 19e^x^2 dx is accurate to within 0.0001?

Question

I'm very lost on the following problem and will appreciate your help very much.

How large should $n$ be to guarantee that the Simpson's Rule approximation on the $\int_0^1 19e^{x^2} \, dx$ is accurate to within $0.0001$?

Answer 1

A bound on the error in using Simpson's rule with $n$ subintervals to approximate the integral of $f(x)$ over $[a,b]$ is

$$E \leq \frac{(b-a)^5}{180n^4}\max_{a \leq x \leq b}|f^{iv}(x)|.$$

Differentiating $f(x) = 19e^{x^2}$ four times we have for $x \in [0,1]$

$$f^{iv}(x) = 19e^{x^2}(12+48x^2+16x^4) \leq 1444e.$$

We have $E < 0.0001$ if

$$ \frac{(b-a)^5}{180n^4}\max_{a \leq x \leq b}|f^{iv}(x)|= \frac{1444e}{180n^4}< 0.0001,$$

Thus $n \geq 22$ is required.

Answer 2

A bound on the error in using Simpson's rule with nn subintervals to approximate the integral of $f(x)$ over $[a,b]$ is

Differentiating $f(x)=19e^{x^2}$ four times we have for $x\in [0,1]$

$$f(x)=38e^{x^2}(8+24x^2+8x^4)$$

plugging $1$ into $x$ we get $$38e^{x^2}(8+24x^2+8x^4) = 1444e$$

and plugging this into the error formula you get

$$\frac{1444e(1-0)^5}{180n^4} \geq .00001$$

calculating for $n$ we get $n \geq 38.14$ and since $n$ has to be even the answer is $n \geq 40$