Q:

Evaluate $\int_{-1}^{0}(x^4-x^2+2)dx$ using Simpson's Rule

We're asked to evaluate the integral $$\int_{-1}^{0}(x^4-x^2+2)dx$$ using Simpson's Rule.

A:

Simpson's Rule states $$Q(f) = \frac{b-a}{6}(f(a) + 4f(\frac{a+b}{2}) +f(b))$$
In our case:
$$a=-1, b=0$$
$$Q(f) = \frac{0-(-1)}{6}(f(-1) + 4f(\frac{-1 + 0}{2}) + f(0))$$
$$\frac{1}{6}(2 + 4\frac{29}{32} + 2) = \frac{2}{3}(1 + \frac{29}{16}) = \frac{15}{8} = 1.875$$
If we evaluate the integral directly, we get $$\frac{28}{15} \approx 1.8667$$ We thus see that our result is "very" close to the exact value.

