**Illustrates**: basic array slicing, functions as first class objects.

In this exercise, you are tasked with implementing the simple trapezoid rule formula for numerical integration. If we want to compute the definite integral

$$ \int_{a}^{b}f(x)dx $$we can partition the integration interval $[a,b]$ into smaller subintervals, and approximate the area under the curve for each subinterval by the area of the trapezoid created by linearly interpolating between the two function values at each end of the subinterval:

<img src="http://upload.wikimedia.org/wikipedia/commons/thumb/d/dd/Trapezoidal_rule_illustration.png/316px-Trapezoidal_rule_illustration.png" /img>

The blue line represents the function $f(x)$ and the red line is the linear interpolation. By subdividing the interval $[a,b]$, the area under $f(x)$ can thus be approximated as the sum of the areas of all the resulting trapezoids.

If we denote by $x_{i}$ ($i=0,\ldots,n,$ with $x_{0}=a$ and $x_{n}=b$) the abscissas where the function is sampled, then

$$ \int_{a}^{b}f(x)dx\approx\frac{1}{2}\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)\left(f(x_{i})+f(x_{i-1})\right). $$The common case of using equally spaced abscissas with spacing $h=(b-a)/n$ reads simply

$$ \int_{a}^{b}f(x)dx\approx\frac{h}{2}\sum_{i=1}^{n}\left(f(x_{i})+f(x_{i-1})\right). $$One frequently receives the function values already precomputed, $y_{i}=f(x_{i}),$ so the equation above becomes

$$ \int_{a}^{b}f(x)dx\approx\frac{1}{2}\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)\left(y_{i}+y_{i-1}\right). $$```
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