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Doss [256]
3 years ago
5

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.) 5 4 1 ln(t) dt, n = 10 (a) the Trapezoidal Rule (b) the Midpoint Rule (c) Simpson's Rule
Mathematics
1 answer:
Otrada [13]3 years ago
4 0

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}

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Let us first make the sample space for the event A. The Sample Space will be:

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(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

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(6,1) (6,2) (6,3)

As can be clearly seen the Sample Space for the event A has 30 elements in it.

Now, it is given in the question that the event A has already occurred and we need to find the probability of the event B occurring <u><em>given that the event A has already occurred.</em></u>

Now, as we can see, from the sample space, only 11 events out of the 30 are the events of interest to us. This is shown in bold below:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

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Thus, the probability that the event B occurs given that the event A has already occurred is:

P(B|A)=\frac{11}{30}\approx0.367

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The lengths of the diagonals of rectangle ABCD intersect at E. If AE =x+4 and CE= 3x-12. What is the length of BD?
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DB = 24

Step-by-step explanation:

First, note that the diagonals of a rectangle are equal and bisect each other. In other words, DB = CA and CE = EA and DE = BE.

Also, AE + CE = CA

So, using this, we can write this equation:

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Subtract 4 from both sides.

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Subtract 3x from both sides.

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Divide both sides by -2

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Then, substitute this into AE + CE = CA

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Then, because CA = DB,

DB = 24

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a) Volume of Rectangular prism is \frac{36}{125}\,\,cm^3

b) Volume of cube is \frac{8}{125}\,\,cm^3

c) Volume of cube is \frac{1}{125}\,\,cm^3

d) Volume of Rectangular prism is \frac{6}{125}\,\,cm^3

Step-by-step explanation:

Part a)

Volume of rectangular prism with

length= \frac{3}{5}\,\,cm

width= \frac{4}{5}\,\,cm

height = \frac{3}{5}\,\,cm

The formula used to find Volume of rectangular prism is:

Volume\,\,of\,\,rectangular\,\,prism=length*width*height

Putting values:

Volume\,\,of\,\,rectangular\,\,prism=\frac{3}{5} *\frac{4}{5} *\frac{3}{5}\\Volume\,\,of\,\,rectangular\,\,prism=\frac{36}{125}\,\,cm^3

So, Volume of Rectangular prism is \frac{36}{125}\,\,cm^3

Part b)

Volume of cube with side length of \frac{2}{5}\,\,cm

The formula used to find Volume of cube is:

Volume\,\,of\,\,cube=length^3

Putting value of length and finding volume:

Volume\,\,of\,\,cube=length^3\\Volume\,\,of\,\,cube=(\frac{2}{5})^3\\ Volume\,\,of\,\,cube=\frac{8}{125}\,\,cm^3

So, Volume of cube is \frac{8}{125}\,\,cm^3

Part c)

Volume of cube with side length of \frac{1}{5}\,\,cm

The formula used to find Volume of cube is:

Volume\,\,of\,\,cube=length^3

Putting value of length and finding volume:

Volume\,\,of\,\,cube=length^3\\Volume\,\,of\,\,cube=(\frac{1}{5})^3\\ Volume\,\,of\,\,cube=\frac{1}{125}\,\,cm^3

So, Volume of cube is \frac{1}{125}\,\,cm^3

Part d)

Volume of rectangular prism with

length= \frac{1}{5}\,\,cm

width= \frac{2}{5}\,\,cm

height = \frac{3}{5}\,\,cm

The formula used to find Volume of rectangular prism is:

Volume\,\,of\,\,rectangular\,\,prism=length*width*height

Putting values:

Volume\,\,of\,\,rectangular\,\,prism=\frac{1}{5} *\frac{2}{5} *\frac{3}{5}\\Volume\,\,of\,\,rectangular\,\,prism=\frac{6}{125}\,\,cm^3

So, Volume of Rectangular prism is \frac{6}{125}\,\,cm^3

Keywords: Volume

Learn more about Volume at:

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