Question

Сalculus2 Please explain in detail if possible

Answer 1

Looks like $n_t$ is the number of subintervals you have to use with the trapezoidal rule, and $n_s$ for Simpson's rule. In the attachments, I take both numbers to be 4 to make drawing simpler.

• For both rules:

Split up the integration interval [1, 8] into n subintervals. Each subinterval then has length (8 - 1)/n = 7/n. This gives us the partition

[1, 1 + 7/n], [1 + 7/n, 1 + 14/n], [1 + 14/n, 1 + 21/n], ..., [1 + 7(n - 1)/n), 8]

The left endpoint of the $i$th interval is given by the arithmetic sequence,

$\ell_i=1+\dfrac{7(i-1)}n$

and the right endpoint is

$r_i=1+\dfrac{7i}n$

both with $1\le i\le n$.

For Simpson's rule, we'll also need to find the midpoints of each subinterval; these are

$m_i=\dfrac{\ell_i+r_i}2=1+\dfrac{7(2i-1)}{2n}$

• Trapezoidal rule:

The area under the curve is approximated by the area of 12 trapezoids. The partition is (roughly)

[1, 1.58], [1.58, 2.17], [2.17, 2.75], [2.75, 3.33], ..., [7.42, 8]

The area $A_i$ of the $i$th trapezoid is equal to

$A_i=\dfrac{f(r_i)+f(\ell_i)}2(r_i-\ell_i)$

Then the area under the curve is approximately

$\displaystyle\int_1^8f(x)\,\mathrm dx\approx\sum_{i=1}^{12}A_i=\frac7{24}\sum_{i=1}^{12}f(\ell_i)+f(r_i)$

You first need to use the graph to estimate each value of $f(\ell_i)$ and $f(r_i)$.

For example, $f(1)\approx2.1$ and $f(1.58)\approx2.2$. So the first subinterval contributes an area of

$A_1=\dfrac{f(1.58)+f(1)}2(1.58-1)=1.25417$

For all 12 subintervals, you should get an approximate total area of about 15.9542.

• Simpson's rule:

Over each subinterval, we interpolate $f(x)$ by a quadratic polynomial that passes through the corresponding endpoints $\ell_i$ and $r_i$ as well as the midpoint $m_i$. With $n=24$, we use the (rough) partition

[1, 1.29], [1.29, 1.58], [1.58, 1.88], [1.88, 2.17], ..., [7.71, 8]

On the $i$th subinterval, we approximate $f(x)$ by

$L_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

(This is known as the Lagrange interpolation formula.)

Then the area over the $i$th subinterval is approximately

$\displaystyle\int_{\ell_i}^{r_i}f(x)\,\mathrm dx\approx\int_{\ell_i}^{r_i}L_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6\left(f(\ell_i)+4f(m_i)+f(r_i)\right)$

As an example, on the first subinterval we have $f(1)\approx2.1$ and $f(1.29)\approx1.9$. The midpoint is roughly $m_1=1.15$, and $f(1.15)\approx2$. Then

$\displaystyle\int_{\ell_1}^{r_1}f(x)\,\mathrm dx\approx\frac{1.29-1}6(2.1+4\cdot2+1.9)=0.58$

Do the same thing for each subinterval, then get the total. I don't have the inclination to figure out the 60+ sampling points' values, so I'll leave that to you. (24 subintervals is a bit excessive)

For part 2, the average rate of change of $f(x)$ between the points D and F is roughly

$\dfrac{f(5.1)-f(2.7)}{5.1-2.7}\approx\dfrac{1.3-2.6}{5.1-2.7}\approx-0.54$

where 5.1 and 2.7 are the x-coordinates of the points F and D, respectively. I'm not entirely sure what the rest of the question is asking for, however...