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+\frac{7(i-1)}n`

and the right endpoint is

`r_i=1+\frac{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=\frac{\ell_i+r_i}2=1+(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=\frac{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=\frac{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)((x-m_i)(x-r_i))/((\ell_i-m_i)(\ell_i-r_i))+f(m_i)((x-\ell_i)(x-r_i))/((m_i-\ell_i)(m_i-r_i))+f(r_i)((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

`(f(5.1)-f(2.7))/(5.1-2.7)\approx(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...