Question

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) 4 e2√t sin(2t) dt, n = 8 0

Answer 1

It looks like the integral could be

$\displaystyle\int_0^4\underbrace{e^{2\sqrt t}\sin(2t)}_{f(t)}\,\mathrm dt$

to be approximated by the three listed rules with $n=8$.

Splitting up the interval of integration into 8 subintervals gives us the partition,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [7/2, 4]

where the left and right endpoints, respectively, are given by the sequences

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

where $1\le i\le8$. The midpoints of each subinterval are given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{\frac{i-1}2+\frac i2}2=\dfrac{2i-1}4$

Each subinterval has length

$\Delta t_i=r_i-\ell_i=\dfrac i2-\dfrac{i-1}2=\dfrac12$

• Trapezoidal rule:

$\displaystyle\sum_{i=1}^8\frac{f(\ell_i)+f(r_i)}2\Delta t_i\approx11.070970$

• Midpoint rule:

$\displaystyle\sum_{i=1}^8f(m_i)\Delta t_i\approx10.767065$

• Simpson's rule:

First we interpolate $f(t)$ over each subinterval with its own quadratic polynomial, given by

$p_i(t)=f(\ell_i)\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

The integral is then approximately equal to

$\displaystyle\sum_{i=1}^8\int_{\ell_i}^{r_i}p_i(t)\,\mathrm dt$

It turns out that

$\displaystyle\int_{\ell_i}^{r_i}p_i(t)\,\mathrm dt=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

so the integral we want to approximate is about

$\displaystyle\sum_{i=1}^8\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))\approx10.868366$

Compare these results to the actual value of the integral, which is about 10.873071.