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adoni [48]
3 years ago
12

Сalculus2 Please explain in detail if possible

Mathematics
1 answer:
Tom [10]3 years ago
8 0

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 <em>n</em> subintervals. Each subinterval then has length (8 - 1)/<em>n</em> = 7/<em>n</em>. This gives us the partition

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

The left endpoint of the ith 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 ith 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 ith 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 ith 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...

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