Simpson's Rule over [-3,3]

To Find $$I = \int_{-3}^3 \sin x^4 dx$$ using the Simpson's (Parabolic) rule with $$n=6$$ intervals.

$$h=\frac{b-a}{n}=\frac{3-(-3)}{6}=1$$

$$\begin{array}{|c|c|c|}\hline x\rightarrow&x_0&x_1&x_2&x_3&x_4&x_5&x_6\\ \hline&-3&-2&-1&0&1&2&3\\ \hline y=\sin x^4\rightarrow&-0.629&-0.287&0.841&0&0.841&-0.287&-0.629\\ \hline &y_0&y_1&y_2&y_3&y_4&y_5&y_6\\ \hline \end{array}$$

According to the Simpson's rule:

$$I = \frac{h}{3}[y_0+y_6+2(y_1+y_3+y_5)+4(y_2+y_4)]$$

$$I = \frac{1}{3}[-0.629-0.629)+2(-0.287+0-0.287)+4(0.841+0.841)] = -0.0657$$

So the result with the Simpson's method is -0.0657. But the actual value of the integral is 0.67946. The result with wolframalpha is 0.294673.

Do we need to break the interval at x=0 in [-3,0] and [0,3] and add the two results?

• This function oscillates wildly (draw the graph!). With so few intervals, one cannot expect an accurate estimate to the integral. – Lord Shark the Unknown Oct 24 '18 at 6:41
• Yes, I see that it oscillates wildly. So can we guess the approximate value of n (number of intervals) for a good result? – simajinid Oct 24 '18 at 6:48

$$-\dfrac{1}{90}\left(\dfrac{b-a}{2}\right)^5 f^{(4)}(\xi) =-2.7 f^{(4)}(\xi)$$
for some $$\xi \in [a, b]$$.
$$\dfrac{d^4}{dx^4}(\sin x^4) = 16(16 x^8 - 51) x^4\sin(x^4) + 8(3-144x^8)\cos(x^4)$$
So, at $$\xi=2.9$$ (for example), $$-2.7 f^{(4)}(\xi) = -2.4484×10^8$$