Use both trapezoid and Simpson's rule to find <msubsup> &#x222B;<!-- ∫ --> 0 <mi

Use both trapezoid and Simpson's rule to find ${\int }_{0}^{\mathrm{\infty }}{e}^{-x}dx$ starting with $h=2$ where $h$ is the length of subintervals $\left[{x}_{i},{x}_{i+1}\right]$.
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sniokd
A well-known trick is to make the change of variables
$x=\frac{t}{1-t}$
So that the integral becomes
${\int }_{0}^{+\mathrm{\infty }}{e}^{-x}\mathrm{d}x={\int }_{0}^{1}\frac{{e}^{-t/\left(1-t\right)}}{\left(1-t{\right)}^{2}}\mathrm{d}t$
the trick here is that this argument goes to zero when $t\to 1$, so you can manually set the last point in your quadrature to $0$