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lapo4ka [179]
2 years ago
5

Approximate the area under the curve y = x² from x = 2 to x = 5 using a Right Endpoint approximation with 6 subdivisions.

Mathematics
1 answer:
Tanzania [10]2 years ago
3 0

Answer:

\text{Area}\,=36.75

Step-by-step explanation:

Using right estimation point simply means to form a bunch of rectangles between the two limits, x =2 and x = 5. and add the areas of all those rectangles.

There must be 6 subdivisions between 2 and 5. so, to do that:

\Delta{x}=\dfrac{5-2}{6}=0.5

the length of each subdivision is 0.5 units. That also means that the 6 rectangles in between the limits will each have the base length of 0.5 units.

So the endpoints of each subdivision from 3 to 5 will be:

\begin{tabular}{|c|c|c|c|c|}3&3.5&4&4.5&5\\\end{tabular}

By <em>right </em>endpoint approx<em>, </em>we mean that the height of the rectangles will be determined by the right endpoint of each subdivision, that is, it must be equal to the function value of the first limit.

\begin{tabular}{|c|c|c|}subdivision&$x$&height($y=x^2$)&3 to 3.5&3.5&12.25&3.5 to 4&4&16&4 to 4.5&4.5&20.25&4.5 to 5&5&25\end

Note that we have used the right-end-point of the subdivision to determine the height the rectangles.

All that's left to do now is to simply calculate the areas of the each of the rectangles. And add them up.

the base of each of the rectangle is \Delta{x}=0.5

and the height is determined in the table above.

\text{Area}\,=(0.5\times12.25)+(0.5\times16)+(0.5\times20.25)+(0.5\times25)

\text{Area}\,=0.5(12.25+16+20.25+25)

\text{Area}\,=36.75

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Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

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The right choice is:

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Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

Its wavelength \lambda_1 = 480 cm

                         \lambda_2 = 160 cm and \lambda_3 = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

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  • The fundamental wavelength of a pipe which is open at both ends is 2L.
  • The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ 4L=4(120)=480\ cm

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒  \lambda_2=\frac{4L}{3}                                   ⇒  \lambda_3=\frac{4L}{5}

⇒ \lambda_2=\frac{4(120)}{3}                              ⇒  \lambda_3=\frac{4(120)}{5}

⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

So the pipe is open at one end and closed at one end .

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