542 L = kL help plzzzzzz

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

4 0

4 years ago

How do you simplify x^1/2 x^3/10 x^2/5/(x^2)^-1/2

shusha [124]

Answer:

the chinese microscpotic flu contains 1 ounce of our blood

jonony me superhero spidermam

Step-by-step explanation:

jonony me superhero spidermam

7 0

3 years ago

Joe's new car travel 418 miles on 16 gallons of gas find the unit rate in miles per gallon show your work I will Mark brainiest

seraphim [82]

To find the unit rate, divide total miles by number of gallons:

418 miles / 16 gallons = 26.125 miles per gallon
Round the answer as needed.

5 0

3 years ago

The cost of 5 diesels is $1025. Calculate the cost of 17diesels

Talja [164]

To calculate the cost of 1 diesel divide the cost of 5 by 5 to get 1 or $1025/5=205
To get the cost of 17, multiply the cost of 1 which is 205x17=3,485 so 17 is $3,485

6 0

3 years ago

Can someone help me out?

Lera25 [3.4K]

Answer:

so basically

the area of the square; (5cm)^2 = 25cm^5

the area of the triangle; (4cm.5cm)/2 = 10cm^2

which means that r = √25cm^2/10cm^2 = √25/10 cm^2 or √10/25 cm^2

english is not my first language so i do not know what combined means but if u explain a little more i will try to help u

-> ok edit edit;

1cm/5in. = r

5cm => 25in.

z^2= area of a square so

25in.25in= 625 in^2

4cm => 20in.

5cm => 25in.

h.b/2 = area of a triangle so

20in.25in./2 = 250in.^2

250in.^2 + 625in.^2 = 875in.^2

so answer B

5 0

3 years ago

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