Question

A region in the plane is bounded by the graph of y=1/x, the x-axis, the line x=m and the line x=3m, m>0. the area of this region:
a) is independent of m
b) increases as m increases
c) decreases as m increases
d) decreases for all m<1/3
e) increases for all m<1/3

Answer 1

Answer:

The area of plane is independent of m, A=log 3

A is correct

Step-by-step explanation:

A region in the plane is bounded by the graph of

$y=\frac{1}{x}$

x-axis

Line: x=m and x=3m

Please find the attachment for figure.

Area under the curve = integrate the curve from m to 3m

$A=\int_{m}{3m}\frac{1}{x}dx$

$A=\log x|_{m}^{3m}$

$A=\log 3m-\log m$

Using log property: log a - log b = log(a/b)

$A=\log(\frac{3m}{m}$

$A=\log 3$

Hence, The area of plane is independent of m.

Answer 2

The area bounded by the graph of $\rm y=\frac{1}{x}$ , the x-axis, the line x=m and x=3m is log3, which is independent of m ie. option (a) is true.

It is given that the region in the plane is bounded by the graph of  $\rm y=\frac{1}{x}$ ,

the x-axis, the line x = m and x = 2m where m >0.

It is required to find which of the following statement is true.

What is the area under the curve?

It is defined as the area bounded by the curves it can be calculated using definite integrals between the two limits ie. lower and higher limits.

We have curves such that:

$\rm y=\frac{1}{x}$ , x-axis , the lines x = m, and x =2m

The area under the curve = (integration of the given curve between m to

                                              2m)      

$\rm A= \int_{m}^{3m}\frac{1}{x}dx\\\\\rm A = [logx]_{m}^{3m}\\\\\rm A = log3m-logm\\\\\rm A = log\frac{3m}{m} \\\\\rm A = log3$∵ $\rm [{\int \frac{1}{x} dx = logx }]$  and  $\rm[logx-logy=log\frac{x}{y} ]$

Thus, the area bounded by the graph of $\rm y=\frac{1}{x}$ , the x-axis, the line x=m and x=3m is log3, which is independent of m ie. option (a) is true.

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